Mastering Sequence and Series: A Comprehensive Guide

today we'll study sequence and series and this first question is let A1 A2 up to a n + 1 b and a and let its

common difference be D now it has defined S1 as some of the series from A1 to a n S2 is from a n + 1 to a2n S3 is

from a2n + 1 to a3n so each of the series contains n terms and S1 is sum of first end terms S2 is sum of next end

terms S3 is sum of again next end terms now it says prove that the sequence S1 S2 S3 it is an AP whose common

difference is n² * the common difference of given progression now for this what we'll do

is we'll write this General series SK and we need to find the first ter of the series now

if it is S1 last term is a n for S2 it is a2n for S3 it is a3n so for s k 1 its last term will be

a k minus 1 n now here its first term will be a k minus 1 n + 1 and then we'll have n

terms and it's last term will be a k n now this is sum of N terms of an AP so

it sum will be n by 2 2 A and A here is a K -1 n + 1

+ n -1 into D and then we can write SK + 1 now here first term will be a k n + 1 and go all the way

a k + 1 n and this sum it be n by 2 2 a k n + 1 + n -1 into D now this C quence it is an AP then difference between any

two successive terms must be independent of K so what we'll do is we'll find S + 1 minus SK and here n minus 1 D

will cancel so we can write this as n by 2 and this is 2 a k n + 1 - 2

a k -1 n + 1 now this 2 and two it also cancel so this is n now this is k n + 1 term so it'll be this A1

+ K N into D and this is minus and this is A1 + K -1 n into D now this A1 and A1 will

cancel and this K and D will also cancel so this value will be equal to n² D which is independent of K now since it

is independent of K it means this sequence it is an AP and this common difference it is equal to n² into D

where D is the common difference of the given progression and which is what we need to prove in this question now here

the question is let A1 A2 A3 a n b and a then we need to prove that 1 upon A1 a n plus 1 upon A2 a nus1 + 1 upon A3 aus2

up to 1 Upon A N A1 is equal to this right hand side now we have this one interesting property in arithmetic

progression and the property is arithmetic mean of terms which are equidistant from beginning and

the end are equal so that means A1 + a n by 2 it is equal to A2 + a n - 1 by 2 it is equal to A3 + a nus

2x2 and L continue like this now we can cancel this two also so we can say some of these numbers they'll be equal now

we'll take the series which is 1 upon A1 a n + 1 upon A2 a n minus 1 1 upon A3 a n minus 2 up to 1 Upon A N A1 and what

we'll do is we'll multiply and divide everything with A1 + a n so we'll write 1

upon A1 + a n and we'll multiply everything with it so it is A1 + a n upon A1 n

+ A1 + a n upon A2 a n minus 1 A1 + a n

upon A3 a n minus 2 all the way up to A1 + a n Upon A N A1 now we can write this as 1

upon A1 + a n and this is A1 + a n upon A1 a n now we know that A1 + a n is also equal to A2 + a n minus1 so we'll

replace it here so we'll write A2 + a nus1 upon A2 a n minus 1 in the same way A1 + n is also equal to A3 + a nus 2 so

it will be A3 + a nus 2 and this is A3 a nus 2 I'll continue up to A1 + a n Upon A N A 1 now what

we'll do is we'll split these terms so we can write 1 upon A1 + a n and here will will be 1

upon a n + 1 upon A1 and here will be 1 Upon A N - 1 + 1 upon A2 1 Upon A nus 2 plus 1 upon A3 it continue up to 1 upon

a n + 1 upon A1 so here we have we have this series 1 upon A1 + 1 upon A2 + 1 upon A3 up to 1

upon a n and again we'll have the series A1 A2 up to a n minus 2 a nus1 a and so we have the series twice so we can

actually write this as 1 upon A1 + a n and it will be 2

* 1 upon A1 + 1 upon A2 up to 1 Upon A N so this is equal to 2 upon A1 + a n into 1 upon A1 + 1 upon A2

up to 1 Upon A N which is what we need to prove in this question now the second question is show that in an AP this

value it is equal to right hand side now we have this AP A1 A2 A3 and let its common difference be D now we are given

the series as A1 s - A2 s + A3 s - A4 squ and it goes all the way up to

a 2 K - 1 s- a 2 k s now what we'll do is we'll pair these terms taking two at a

time so this is a squ - b square so we can write this as a + b into a minus B so there'll be A1 + A2 and into A1 - A2

and this is A3 + A4 into A3 - A I'll continue up to a 2K - 1 + a 2K into A2 K -1

minus A2 K now we know that in an AP difference between any two successive terms it is either D or minus D since it

is A1 - A2 so it'll be this minus d will also be minus D and in the same way it is also minus D so we'll take this minus

D common and then we'll have A1 + A2 + A3 + A4 up to a 2K which is sum of first 2 G terms of

an AP now this is minus D and this is n by2 so it'll be 2 K by 2 and a + L which is A1 + a 2 K now this 2 and two will

cancel so it is equal to minus D into K this is A1 + a

2K now we also know that in an AP this a2k will be equal to A1 + 2 K - 1 into D so we can write this D

is A2 K - A1 upon 2 K - 1 so we'll put it here so if we put it here we'll get

minus and this is A2 K - A1 upon 2 K - 1 into K into A1 + A2 K now we'll take this minus sign here then

will be a + b into a minus B which is K upon 2 K -1 into A1 s

minus A2 k s which is what we need to prove in this question now in this question it says if

the roots of the equation 10 x CU - CX2 - 54x - 27 = 0 are in HP then find C and find all the

roots now it is easier to work with an AP rather than an HP so what we'll do is we'll rather find an

equation whose roots are reciprocal of roots of given

equation so for this what we simply do is we replace x with 1X X so we can write

10 1X X Cub - c 1X x² - 54 1X x - 27 = 0 now if we simplify and rearrange we'll get this as 27 x

Cub + 54 x² + CX - 10 = 0 now since its roots are

reciprocal to the roots of given equation roots of this equation they are in a and let these roots

be a minus d a and a + D now we can write S1 S1 is sum of all the three Roots

which in this case is 3 a and that'll be equal to- B Upon A so it'll be - 54 upon 27 which is -2 so value of a is

minus 2x3 so we have this a as Min - 2x3 now since this a is root of this equation it'll satisfy this equation so

we'll put the value of x and find the value of C so that'll 27 and this is - 8X

27 + 54 and this is 4X 9 - 2 C by 3 - 10 = 0 now this is - 8 + 24 - 2 C by 3 - 10 = 0 now this

is 2 C by 3 = 6 value of C is 9 so we'll get the value of C is 9 so in this question value of C

is 9 now we need to find the other roots and for it we need to find the value of D so what we'll do is we'll write

S3 now S3 is Alpha Beta gamma so it'll be a into a s - D ² and that'll be equal to minus D Upon A which

is 10 by 27 the value of a is - 2x3 so there'll be - 2x3 and this is 4X 9 - D ² and this is 10 by

27 now this is 5 and this is 9 so we can write -4 + 9 D ² = 5 so value of D is Plus or

-1 Now The Roots are a a minus D and A + D so roots of this equation there will be a minus D so there'll be - 2 by 3 - 1

so there'll be - 5x3 and then a and then a + D which is 1x 3 but these are roots of the equation whose roots are

reciprocal to the roots of given equation so we need to find roots of the given

equation so roots of given equation they'll be reciprocal to these roots so there'll be - 3x

5 - 3x2 and3 and the value of C we'll have here is 9 and that is the answer to this question now here the question is

if the sum of M terms of an AP is equal to sum of either the next n terms or the next p terms then prove that 1 upon m +

n into 1 upon M - 1 upon P will be equal to m + P into 1 upon M - 1 upon M where n is unequal to P so what we given is

sum of M terms is equal to sum of next n terms now we can write sum of next n

terms as sum of m+ N minus sum of N terms so it will give us the sum of next n terms so we can write 2sm will be

equal to s m + n and we have given that it is also

equal to sum of next p terms so another condition which we'll have here is 2m will be equal to

s m + P where n is unal to P now we'll simplify this first conditions we can write two and this is M by

2 2 A + m-1 into D and it'll be equal to m + n/ 2

2 a + m + N - 1 into D now this 2 and two will

cancel now what we'll also do is we let this 2 a + m-1 DS sum X so we can write 2 MX will be equal

to m + n and here will be this 2 a + m-1 D is X plus n into

D now we'll take this X on the left hand side we can write M - n into X and that'll be equal

to m + n into n d or we can simply write X upon D will be equal to n

into m + n upon M minus n now similarly we can write it for this equation and from here we'll

get this x upon DS P into m + P upon M minus P now we equate these two

we'll get n into m + n

upon nus n and that be equal to P into m + P upon m-

V now we can rearrange this as m + n into m-

P into n and that'll Beal P into m + P into M minus n now we'll divide everything

with M and and we'll divide it with these two terms so we'll

write m + n now this n will cancel and will be 1 upon P minus 1 upon M and here will be this m + p and it will be 1 upon

n minus 1 upon n which is what we need to prove in this question the question is

computer solved several problems in succession the time it took the computer to solve each successive problem was

same number of times smaller than the time it took to solve the preceding problem how many problems were suggested

to the computer if it spent 63.5 minutes to solve all the problems except for the first and 127 minutes to solve all the

problems except the last one and 31.5 minutes to solve all the problems except the first two now let the time taken for

the first problem be a for the second it is a then it is AR squ a RQ and go all the way up to a r the^ n -2 and a r^ n

minus one so in total you'll have n problems now it says it spent 63.5 minutes to solve all the problems except

for first so for all these problems it has taken the time

63.5 127 minutes to solve all the problems except the last one so for all the problems except the last one it has

127 minutes and 31.5 minutes to solve all the problems except for first two except

for first two time taken is 31 5 minutes now if we write this sum

from a to a r^ n minus 2 this sum will be a 1 - r^ n -1 upon

1 - R it is 127 and this sum from a r to a r^ n minus one there'll be a r 1 - r^ n - 1 upon 1 - R and it will be

63.5 if we divide first with second everything will cancel you'll get this as 1 upon R and that'll be equal

to 127 upon 63.5 which is two so value of R is

[Music] 1x2 now we know that time taken to solve problems from 2 to last it is 63.5 and

time taken to solve problems from 3 to last is 31.5 that means time taken to solve the second

problem it is a and it is equal to 63.5 - 31.5 so it will be 63.5 minus 31.5 and that is 32 now if you put R is

1x2 you'll get value of a is 64 so value of a is 64 so we have a and r we need to find the value of n so we put the value

of a and r in this equation so now we can write

64 1 - 1 upon 2^ N - 1 upon 1 - 1 by 2 and that'll be equal to 127 so this is 128 so we can write 1 - 1 upon 2^ n -1

will be equal to 127 upon 128 or 1 upon 2^ N - 1 is 1 upon 128 now this is 1 upon 2 ^ 7 so N - 1 = 2^ 7 so value of n

is 8 so in total eight problems were given to the computer and that is the answer to this question now question is

a GP and HP have same p q and R term ABC show that a into B minus C log a plus b into c - a log B plus C into Aus B log C

it is equal to 0 and it says a BC are p q and R term of an HP so that means 1 Upon A 1 upon B and 1 upon C are p q and

R term of for an AP now we can write this as A+ a - 1 D it is equal to 1 by a a + Q - 1

D it is 1 by B and A + r - 1 into D is 1 by C now if we subtract these two we can write P - Q into D it is equal to 1 upon

a - 1 upon B and in the same way if we subtract these two we can write Q - Rd = 1 upon b -

1 upon C now we are also given that

abcr p qth and R term for a GP so we can write this a some a

Das r^ P minus one B is some a Das r^ q minus1 and C is some a d r the^ r - one and if we take log we can write log a

equal log a d + P minus1 log R log B and this log a d + Q -1 into log R and log C = log a d

+ r -1 into log we'll take everything as log now if we subtract these two we can write log a minus log B it is equal p- Q

into log R and from second and third we can write log B

minus log CS Q - R into log R now we'll come back to this

first one if we divide these two we can write p- Q upon Q - R it will be equal to B minus a upon AB

into C minus B to BC which is B minus a c Upon A into cus B and if

we divide these two we can write log a minus log B upon log B

minus log c will be equal to p- Q upon Q - R now if we equate these two we can

write C into B- a Upon A into cus B will be equal to log a minus log B

upon log B minus log C now we cross multiply we can write

C B minus a log B minus C B minus a log c will be equal to a c minus

B log a minus a c minus B log B now we'll take everything on the

left hand side then we can write this as a B minus C log a now plus and we take

log B on the left hand side it'll be BC minus AC and AC minus AB so there'll be b c minus

a log B and this is plus we'll take this plus sign so it will be C into a minus B log C = 0 which is what we need to prove

in this question now here it says let S and B sum of nend terms of a GP whose first

term and common ratios are A and R then do that S1 + S3 + S5 to S 2 N - 1 is a n upon 1 - r - a r 1 - r^ 2 N upon 1 - r²

into 1 + r now we know that SN is a into 1 - r^ n upon 1 - R now we need to find

the value of S1 + S3 + S5 up to S 2 nus1 now s 1 is a 1 - R upon 1 - R S3 is a into 1 - R

upon 1 - R S5 is a 1 - r^ 5 upon 1 - r n s 2 N - 1 will be a 1 - r ^ 2 N - 1 upon 1 - R

now we'll add all all these A1 - R terms from all the Expressions now number of terms in the

series is n so it'll be a n upon 1 - R and then we left with this minus we'll take a r upon 1 - R

common and then we'll have 1 + r² + r ^ 4 up to r^ 2 N minus 2 which is nothing but a GP

so we can write this as a n upon 1 - R minus a r upon 1 - R and this is a 1

- r² ^ n upon 1 - r² now this is a n upon 1 - R and this is minus a r

upon 1 - R and this is 1 - r ^ 2 N and this is A + B into a minus B so it'll be 1 - R into 1 + r so this

sum it will be a n upon 1 - r - A R into 1 - r^ 2 N upon 1 - r² into 1 + r which is what we need to

prove in this question now other question is Let There Be Maps beginning with unity so for each

of the aps first term is one and their common differences are 1 2 3 up to M so their common differences are I show that

some of their nth term is M by2 into MN minus m + n + 1 so if we write nth term for this ith series it will be a i

+ nus 1 into I so will be 1 + nus1 into I now we need to find sum of

all these terms where I varies from 1 to M now it'll be this summation I varies from 1 to m 1

+ N - 1 into I now summation 1 is m and this is nus1 and summation I is M into m + 1 by 2 now what we'll do is we'll take

M by2 common so if we take M by2 common you'll get this as 2 + n -1 into m + 1 and this is M by 2 and

here will be this MN + n- - M + 1 which is what we need to prove in this question now here the

question is the sum of squares of three distinct real numbers which are in GP is s squ and their sum is alpha s show that

Alpha Square belongs to 1x 3 to1 Union 1 2 3 so we have three numbers in GP and let the numbers be a a r and a r squ we

are given a + a r + a r² and this is equal to alph and some of their s a² into 1 + r² + r ^ 4 and this is equal to

s² now what we'll do is We'll add R square and we'll subtract R square so we can write this as a square and here will

be 1 + r² 2 - r² and this is equal to s² now this is A + B into a minus B so we can write this

as a 1 + r + r² into a 1 - r + r² and this is equal to s² now we know that a 1 + r + r² it is nothing but it is Al so

we'll replace it with Al s will cancel once so we can write a 1 - r + r² and that'll be equal to S

upon Alpha so this is our first this is second and this is our third result now what we'll do is

we'll divide first with third then we can write 1 + r + r² upon 1 - r + r² and that'll be

equal to Alpha squ now we need to find range of alpha Square now we can write this

as Alpha squ - 1 into r² minus and then it will be alpha s + 1 into r

+ Alp s - 1 = 0 now we need a real value of R so for this D should be greater than zero we

are not considering D equal 0 because if D is equal to Z then both the roots they'll be equal R1 = R2 and we know

that product of root is + one that means R1 and R2 will take the value either + one or minus one in which case number

won't be distinct so we are discarding this case so if we take D than 0 we'll get Al s + 1 s - 4 Alp s - 1 s and it

should be greater than Z now we write a + b into a minus B so it'll be this 3 Alp s - 1 into 3 - Alp Square it should

be greater than Z now if we write for Alpha Square this is 1 by 3 this is 3 3 this is minus plus and minus what we

need is plus so value of alpha Square should lie between 1X 3 and 3 now we also know that R cannot be zero because

again if R is zero then we'll get two terms as zero in which case we'll get Alpha Square = 1 so

R cannot be zero that means value of alpha Square it cannot be one so from this we'll remove one value which is one

so answer to this question is Alpha Square should lie between 1X 3 to 1

Union 1 to 3 and that is the answer to this question now the question is a series of

natural numbers is divided into groups in the first group we have one in the second group we have 2 3 4 in the third

group we have 5 6 7 8 and 9 and so on we need to find some of the numbers in the nth group so we are talking about

10th group now in first group we have one number in second group we have three numbers in third group we have five

numbers so in nth group so in nth group we'll have a + n -1 into D that is 2 N - 1 numbers

we'll have 2 N minus one numbers now we know that in each

group numbers are in AP whose common difference is one so we have common difference of one we need to find this

first term now there are many ways to find this first term now what we can see is here this is one for the second one

it is two for this third one it is nine so this is so this is one square this is 2 square for the third it is 3 Square

for this n minus one group last number will be n -1 squ so here this first term it

will be n minus 1 + one term so for this nth group

is n -1 s + 1 common difference is one and number of terms it is 2 nus 1 we just need to find its sum

so some of these numbers will be n by 2 so it'll be 2 N - 1 by 2 and then a + L so it'll be this N - 1

s + 1 plus and we know that here last term will be n² so it be simply n² so we can write this

as 2 N - 1 by 2 and here will be 2 n² - 2 n + 2 which will cancel so it'll be this

2 N -1 into n² - n + 1 now we have to express it in the form n -1 Cub + n Cub so what we'll do is we write this as n

+ n minus1 and here we'll subtract n and add n so there'll be n -1 s + n now we multiply them we can write n into n -1 2

+ n² + n -1 Q + n into nus1 so it'll be this NB + N - 2 n 2 + n 2 + n -1 CU + n²

- n now this n sare n square and 2 N Square will cancel n and n will also cancel so it'll be this n -1 CU

+ n CU and that is what we need to prove in this question now for this question another

way to find this first term is now in this first group there is only one term so this next term is this 1 + 2 now

here we have 1+ 3 four terms and this is 4 + 1 fifth term now here you have 1 + 3 + 5 9 and here will be this 10 so this

first term it will be 1 + 3 + 5 up to now till n minus 1 group so you'll

have n minus one terms + one now we know that sum of first n odd natural numbers is n Square here we have nus1 terms so

there'll be n -1 Square + 1 so that is another way of finding this first term which again is n -1 s + 1 now here the

questions we are given x y and z now X is this GP 1 + cos² 5 + cos ^ 4 Pi up to infinite so

this is a upon 1 - R which is 1 upon sin Square 5 now this Y is 1

+ sin S 5 + sin ^ 4 5 up to infinite and this is 1 upon 1 - sin Square 5 which is 1

upon cos² 5 and Z is 1 + cos² 5 into sin S 5 + cos ^ 4 5 into s^ 4 5 up to infinite which again is an

infinite GP which is a upon 1 - R now what we'll do is we'll put the value of cos Square 5 and sin Square 5

so from here we can write sin Square 5 is 1x X and cos squ 5 as 1 by y so from here we

can write this Z it is equal to 1 upon 1 - 1 upon XY or Z is equal XY

upon XY - so from here we can write this XY Z it is equal to XY + which is what we need

to prove in this first part now for the second part we need to prove that XY = x + y now we know that sin sare 5 + cos

sare 5 is 1 so if we add them we can write sin Square Theta + cos sare thet 1 and that'll be equal to 1 by x + 1 by y

that means X into Y = X + Y so we'll replace this x into y here so from here we'll get x y z = x + y + z now here

this first question is we are need to find solve the series which is summation R varies from 1 to n summation s varies

from 1 to n Delta RS and this is 2^ R into 3 ^ s where Delta RS is zero if R isal to s and this is one if R is equal

to S so though it's a double summation problem problem is very simple so when R and S they're different this entire term

will be zero so we'll only have non-zero terms when R equals s so that means this s it will be when R is 1 and S is 1

it'll be 2 into 3 when R is 2 s is2 there'll be 2 s into 3

S 2 cu into 3 CU and continue up to 2^ n into 3^ n it is nothing but just another way

of writing a GP so this is another notation in which we can write a GP so this s is nothing but a GP which is 6 6

s 6 Cub up to 6^ n which is a r^ n-1 upon R -1 and that is the answer to this question

now for this second part it says you need to find this triple sum I varies from 1 to

n j varies from 1 to I and K varies from 1 2 J and this is

one now first we'll solve this now this is 1 + 1 + 1 J * so it will be this summation I varies from 1 to n and this

is summation J varies from 1 to I and this is J now summation J is I into I + 1 by 2 so we can write this as this

summation I vares from 1 to n i into I + 1 by 2 or this is this summation i²

+ summ I / 2 now summation i s is n n + 1 2 n + 1 by 6 and summation I is n into n

+ 1X 2 now what we'll do is we'll take n n + 1X 6 common we can write this

as n n + 1 by 12 and here we'll have 2 n + 1 and here will be

simply 3 so this is n n + 1 and this is 2 n + 4 which is 2 into n + 2 upon 12 which we can also

write as n n + 1 n + 2 / 6 and that is the answer to

this question now here the question is we need to find some of these four series

now this first one is we given the series which is 1 into 4 into

7 1 upon 4 into 7 7 into 10 1 upon 7 into 10 into 13 we need to find its

sum up to nth term and then up to Infinity now first we'll find nth term for this series now this is 1 147 it's

an AP so its nth term will be a + n minus one and common difference is three so that'll

be 3 N - 2 so it'll be this 1 upon 3 N - 2 3 n +

1 and 3 n + 4 now basically this question is method of difference now difference between 1 and 7 is 6 4 and 10

is 6 7 and 13 is 6 and 3 N - 2 and 3 n + 4 is 6 so what we'll do is we'll multiply and divide everything with six

so we can write SS 1X 6 and this is 6 upon 147 and this is 6 upon

4710 up to 6 upon 3 N - 2 3 n + 1 3 n +

4 now this is 1 by 6 now we'll write this 6 as 7 - 1 so it will be this 7 - 1 and this is 147 here we'll write this as

10 - 4 this is 4 into 7 into 10 here it is 13 - 7 this is 7 into 10 into

13 and here it is 3 n + 4 - 3 N - 2 so it will be 3 n + 4 - 3 N - 2

[Music] upon 3 N - 2 3 n + 1 3 n + 4 and what we'll do is we'll

split these terms so we can write this is 1x 6 now here 7 will cancel so it is 14 -

47 now here 10 will cancel it will be 4 into 7 and here 4 will cancel 7 into 10 here 13 will cancel so this is 7 into 10

minus 10 into 13 and continue and here this 3 n + 4 will cancel so you'll be this 3 N -

2 3 n + 1 minus and here 3 N - 2 will cancel so it will be 3 n + 1 and 3 n + 4 now if we look at

these terms these successive terms they'll cancel now here we have this first one then what remains is this last

term so this sum it will be 1X 6 into 1X 4 - 1 upon 3 n + 1 into 3 n + 4 and that is

the sum up to end terms now if we find sum up to infinite we simply need to take this limit n TS

to infinite SN now if we take this limit n TS to infinite this entire expression will be simply zero so some of this

limit up to Infinity is 1x 24 so that is the answer to this first part now this second question is find

the sum of the series up to end terms so this series s is summation R varies from 1 to n

r r + 1 r + 2 r + 3 now one way of solving this question is by taking this product and

then we'll have r^ 4 RQ which is very lengthy way of solving this question so what we'll do is we'll write its R term

as r r + 1 r + 2 and r + 3 now we'll try and use method of difference in this so the term before this one it will be r -

1 and term after r + 3 will be r + 4 and the difference is five so what we'll do is we'll multiply and divide everything

with five so we can write 1X 5 and this is r r + 1 r + 2 r + 3 into 5 now we want to write this as 1X 5 and this is

r r + 1 r + 2 r + 3 and we write this as r + 4 4- R-1 and we'll split it in two parts we

can write 1X 4 and this is r r + 1 r + 2 r + 3 r + 4 and minus and this is R -1

r r + 1 r + 2 r + 3 so that's our R term now we can write this first term as 1X 5 and this is 1 into 2 into 3 into 4 into

5 minus and this is 0 we can write its second term as 1X 5 and this is 2 into 3 into 4 into 5 into

6 minus and this is 1 into 2 into 3 into 4 into 5 third term is 1x 5 and this is

3 4 5 6 7 minus and this is 2 3 4 5 and six and its nth term is 1x 5 n n +

1 n + 2 n + 3 n + 4 minus n -1 n n + 1 n + 2 n + 3 now SN is sum of these first end terms so we'll add them we can

write this SN as T1 + T2 plus T3 up to TN now here these terms they'll cancel it cancel cancel now here we have this

left so here what remains is this expression so it'll get cancelled so it'll be this 1X 5 into

n n + 1 and + 2 n + 3 into n + 4 and that is the answer to this question n since SN is

proportional to n s infinite will be infinite so as n approaches infinite this sum is going to

diverge now this third one is this summation R varies from 1 to n 1 upon 4 r² - 1 so it will be this

summation R from 1 to n and this is 2 R -1 and this is 2 r + one now here difference is two so what

we'll do is we'll multiply and divide everything with two it will be the summation R varies from 1 to n and this

is 2 upon 2 and this is 2 R -1 and 2 r + 1 now we can write this as this summation we'll take 1 by 2 2 Common and

here will be this R varies from 1 to n and we can write this as 2 r + 1 minus 2 r - 1

/ 2 R -1 and 2 r + 1 which is 1 by 2 and then this summation R varies from 1 to n now here 2 r + 1 will cancel so will be

this 1 upon 2 r - 1 and here 2 r - 1 will cancel and this is 2 r + 1 so we can write this sum of N terms as 1X 2

now we put r as 1 it will be 1 upon 1 minus 1X 3 R is 2 so it be 1X 3 - 1X 5 and continue up to 1 upon 2 N - 1 - 1

upon 2 n + 1 now 1X 3 1X 5 they'll cancel so here we have this first term we'll have this last term so this SN it

will be 1X 2 1 - 1 upon 2 n + 1 or this is n upon 2 n + 1 now we find S infinite it be this limit n TS infinite n upon 2

n + 1 and since this is infinite limit answer to this limit is coefficient of highest power of n in the numerator and

in the denominator so it will be this one and two so some of the series up to Infinity is 1 by two and then we have

this fourth one which is this series SN so this is 1x 4 1 into 3 and this is 4 into 6 this is

1 into 3 into 5 and this is 4 into 6 into 8 and goes up to nend terms now for nth term we'll have 1

35 so it is an AP so this n term will be a + nus1 into D so that'll be 2 N minus one and here this is

4 6 8 which again is an AP so this is a + nus1 into D which is 2 n + 2 4 68 and this is 135 what

seems to be missing is two so what we'll do is we'll multiply and divide everything with two so we can write this

SNS 2 into and this is 2 into 4 and this is 1 into 3 and this is 2 into 4 into 6 this is 1 into 3 into 5 and this is 2

into 4 into 6 into 8 and this is 1 into 3 into 5 up to 2 nus 1 and here it will be 2 into 4 into

6 up to 2 N + 2 now if we write its general term its

general term will be R term and will be this 2 into two and then 1 3 5 up

to 2 R -1 and in the denominator will be 2 into 4 into 6 up to 2 r + 2 now because we have to use

method of difference somehow we have to express it as sum a minus B so we have to introduce a minus B into this so what

we'll do is we'll write this as one here and we can write this 2 and this is 1 into 3 into 5 up to 2 r - 1 and we can

this as 2 r + 2 minus 2 r + 1 all divide by 2 into 4 into 6 up

to 2 R and this is 2 r + 2 what we'll do is

we'll split it in two parts now we can write this as Dr and this is two now for this first

part 2 r + 2 will cancel so we'll get this as 1 into 3 into 5 up to 2 R -1 upon 2 into 4 into

6 up to 2 R and then minus here it is 1 into 3 into 5 and this is 2 r + 1

upon 2 into 4 into 6 up to 2 2 r + 2 so that's your TR R now if we write T1 T1 will be 2 and

here it is 1x 2 minus and this is 1 into 3 and 2 into 4 because we put r as one this is two and this is four and here

it'll go up to three now if we put DS2 will be this two now this is 1 3 and this is from 2 to 4 minus and this

is 135 2 46 and it'll continue like this and this

n term will be 2 and this is [Music] 135 2 nus1

upon 246 2 N minus and this is 135 2 n + 1

upon 2 4 6 2 n + 2 now SN is sum of all these end terms so if we add them we'll get SN now

here these successive terms they'll cancel and since we have this term here what will remain is this last so it'll

cancel so it'll be this 2 into 1X 2 minus 1 into 3 into 5 up to 2 N +

1 upon 2 into 4 into 6 up to 2 N + 2 and we find it sum up to infinite it will be this

limit n TS infinite s now it be this 2 into 1x2 minus now here number of terms is n and here number of terms is n + 1

so here in the denominator power of n will be greater so it will be simply zero so answer to the series will be one

and that is the answer to this question now here we are given two questions and for these two questions we have to find

their nth term and sum of end terms now this first question is we have the series 1 + 5 +

13 + 29 + 61 up to end terms now here difference is

4 8 16 32 which is a GP so this difference

it is in GP with common ratio two so we have this common ratio of two and we know that if difference is in GP then

its general term it is written as a + b into r to^ n minus 1 and R in this case is 2 so it'll be this A + B into 2

the^ of n minus one now here we have two unknowns A and B we put n is one it will be

this A + B and it be one and if we put ns2 will be a + 2 B and it is five now we

subtract you'll get b as four and if B is for value of a is -3 so its general term will be -3 + 4 into 2^ of n -1 or

it is - 3 + 2^ of n + 1 now we need to find sum of N terms now this SN will be simply this

summation R varies from 1 to n t r which is this summation R from 1 to

n - 3 + 2^ of r + 1 now summation of Min -3 simply - 3 n and then we'll have this

summation R varies from 1 to n 2^ r + 1 which is nothing but a GP now for this GP this first term is when you put r as

one which is four common ratio is 2 and number of terms is n so we can write this as - 3 n and this is 4

into 2^ N - 1 upon 2 - 1 so it will be this - 3 n + 2^ of n + 2 - 4 and that's the answer to this first

question now this second question is we need to find some of the series and it's nth term 6 13

22 33 now here difference is 7 9 11 now difference of difference is constant we know that it's nth term it

will be a n² + BN + C now we'll use the shortcut this is 2 a this is 3 A + B and this is a + B+ C

now clearly a is one and if we put a as one we'll get b as 4 and if we put a as

one and B as 4 we'll get C as one so its general term or its n term will be n² + 4 n + 1 now we need to find sum of N

terms and there'll be this summation R varies from 1 to n TR R so it'll be this summation R vares from 1 to n and

there'll be r² + 4 r + 1 now summation r² is n n +

1 2 n + 1 by 6 Plus and this is 4 n n + 1 by 2 and then +

n now we'll simplify these two what we'll do is we'll take n n + 1X 6 common so we'll have

have 2 n + 1 + 12 + n so it'll be this n n +

1 2 n + 13 upon 6+ n and that is the answer to this question now the next question is we

need to find the sum of the first n terms of the sequence s which is 1 + 2 into

1 + 1 byn + 3 into 1 + 1 by n² 4 into 1 + 1 byn whole Cub up to and the nth term will be n into 1 + 1 by

n to^ nus1 now it's an AGP and for an AGP what we'll do is we'll write the series again

and we'll multiply the series with with its common ratio which is 1 + 1 by n and we'll displace the series by one

term so here it will be 1 into 1 + 1 byn 2 into 1 + 1 by n² and then in the end we'll have this n

into 1 + 1 by n to^ n now what we'll do is we'll subtract this second series from first so we'll get -1 by n SN and

here it will be 1+ and this is 1 + 1 byn this is 1 + 1 by n² I'll go all the way up to this term which is 1 + 1 by n ^ N

- 1- n into 1 + 1 by n^ n now it is nothing but a GP with first term 1 common ratio 1 + 1 by

n and number of terms n so it will be this minus 1 by n s n will be a and then r ^ n which is 1 + 1 by n^ n -1 upon R

-1 so it'll be 1 + 1 by N - 1 and - n into 1 + 1 by n to the^ n now this 1 and one will cancel so we'll get this as -1

by n SN and here it will be this n 1 + 1 by n^ N - n and then - n into 1

+ 1 by n the^ n now it'll cancel and we'll take this minus n on the right hand side you'll get this SN as n² so

sum of first nend terms of this sequence is n² now here the question is given that a

^ x = b ^ y = c ^ z = d ^ U where AB b c d are in GP then show that x y z u are in HP now what we'll do is we'll take

this as Lambda then if we take log we can write X log a = y log b = z log c = u log d =

log Lambda or log a equal log Lambda upon

X log P = log Lambda upon y log c

equal log Lambda upon Z and log D equal log Lambda upon U now in the question it says

AB c d they are in GP now we know that if ABC are in GP then log a log B log C and log D they

are in AP now we'll put the of log a log B log C and log B so there'll be log Lambda upon

X log Lambda upon y log Lambda upon Z and log Lambda upon it'll also be an AP now we'll divide

everything with log Lambda and we know that if we divide each term with a constant then it'll still remain an AP

so that means this 1 upon x 1 upon y 1 upon Z and 1 upon U it be an AP and reciprocal of an AP is HP so that means

XY Zu U they in HP and which is what we need to prove in this question now the question is find three numbers a b c

between 2 and 18 so we have to find these three numbers a b c lying

between 2 and 18 such that their sum is 25 so that means A + B + C is 25 and then it says

the numbers 2 A and B are consecutive terms of an AP that means 2 a

= 2 + B and then it says BC and 18 are consecutive terms of a GP so that means c² = 18b now what

we'll do is we'll put the value of a in this equation we'll multiply it with two so we'll

get 2 a + 2 b + 2 C = 50 and 2 a is B + 2 so there will be 3 b + 2 C =

48 and we know that 18 B is c² so what we'll do is we'll multiply everything with 6 so it will be this

18b + 12 C and this is equal to 48 into 6 which is 288 now this 18 V is c² so we can

write c² + 12 C minus 288 is equal to 0 now 288 is 6 into 48 which is 6 into 8 into 6 so we can write this

as c² + 24 c - 12 C - 288 = 0 so from here we'll get C + 24 and C - 12 = 0 The only

positive value of C in this case is C = 12 as a BC lie between 2 and 18 now once C is

12 B is c² by 18 so there'll be 12 into 12 / 18 value of B is

8 and once we have B and C B + C is 20 then the value of a is 5 so we'll get a is 5

B is 8 and C is 12 now the question is an AP and an HP have the same first term and the same last term and the same

number of terms prove that product of R term from the beginning in one series and the AR term from the end in the

other is independent of R so suppose we have an AP whose first term is a and whose last term is B

and suppose number of terms is n now we know that its common difference it will be b - A upon nus1 and its R term from

the beginning it will be a + r -1 into D which is a + r [Music]

-1 b - A upon nus1 or simply we can write a n -1

+ b - a r -1 upon nus1 we can also write this as a n -

r + b r -1 upon n -1 so this is are term of an AP from the beginning now if you look at HP

it says it has same first and last term So This Is A and B again and again number of terms is n now we solve HP by

making it an AP so this is 1 by a and this is 1 by B number of terms is still n now it common difference will be 1

upon B minus 1 Upon A upon N - 1 which is a - b Upon A into n minus one now in this we need to find R term from the end

and we know that R term from the end is equal to N - r + 1 term from the beginning so it'll be this 1 by

a plus n- r into D which is a minus

B upon AB n minus one now since this is corresponding term of an AP so term of

HP which is our term from the end will be reciprocal of this and it will be this

a n minus one and here will be this B nus1 + n-

R A minus B now we'll take terms of B together so it will be a n

minus1 upon and here this B it will be N minus1 - n + r so it be this b r -1 and

+ n minus r into a now it says we have to prove that product of these two terms it is independent of R now we multiply

them we'll get TR R and T HR from end it will be

a n - r + b r -1 upon n -1 into and this is a n -1 upon b r -1 + N - R into a now this expression will cancel nus 1 will

also cancel so this product it is equal to AB which is independent of R and which is what we need to prove in this

question now in this first question it says value of x + y z is 15 if a x y z and b are in AP so it says

value of x + y + z it is 15 if a x y z and b there are in a now we know that in an

AP arithmetic mean of terms equidistant from beginning and end it is equal that means a + b by 2 it is equal to x

+ z upon 2 and it is equal to Y so in this this x + z will be equal to a +

b and this Y is A + B by 2 and this is equal to 15 so from here we can write value of a + b is 10 and

that's our first equation now in the second part it says if 1xx + 1 by y + 1 by Z

= 5x3 and if a x y z and B they are in HP Find value

of a and b now if they are in HP then 1 Upon A 1 upon x 1 upon y 1 upon Z and 1 upon B they are in AP that

means again arithmetic mean of terms equidistant from beginning and end will be same so 1 Upon A plus 1 upon B by2

must be equal to 1 upon x + 1 upon Z by 2 and will be equal to 1 upon

y now this is 1 Upon A + 1 upon B and this is 1 Upon A + 1 upon b/ by 2 so it'll be this 3x2

* 1 Upon A + 1 upon B and that will be equal to 5x3 now we can write this as 3x2 a + b Upon A B and this is equal to

5x3 now we already know a + b is 10 so if we put a + b as

10 we can write this as 3x2 this is 10 upon ab and this is equal 5x3 now

I'll cancel we'll get value of a into B is 9 so a into B is

9 and a + b is 10 now 2 equations two unknowns now you put the value of B we can write a

+ 9 upon a = 10 or a² - 10 a + 9 = 0 get a - 1 a - 9 = 0 so either the value of a is 1 or value of a is 9 and if a is 1

value of B is 9 and if a is 9 value of B is one and this is the answer to this question now this B question is

the value of XY Z it is 15x2 if a XY z and b they are in

AP and this XY Z is 18x 5 if a XY z and b they are in

HP we have to find the value of a and b as assuming them to be positive integers now we'll take this first case

when a x y z and b there in AP now common difference of this AP will be simply B minus a upon

4 so this x will be simply a + D so it will be this A + B minus a by 4 which is

3 A + B by 4 Y is x +

D so be 2 A + B by 4 and Z will be y + D which

is a + 3 B by 4 so we multiply x y and z you'll get 3 A +

B into 2 * A + B into a + 3

B upon 64 it is equal to 15x 2 and this divide by 32 now it says if

a x y z and b they are in HP then X into Y into Z it is 18 by 5 now if a XY Z and be there in

HP then 1 Upon A 1 upon x 1 upon Y and 1 upon Z so it'll be same as this first

equation where a XY z b they are in AP the only difference is here a is 1 Upon A B is 1 upon b x is 1 upon X Y is 1

upon Y and Z is 1 upon Z so here we'll get 1 upon X into 1 upon Y into 1 upon Z as

this 3 Upon A + 1 upon b 1 Upon A + 1 upon B and 1 Upon A + 3 upon

B by 32 and it will be equal to 5 by 8 18 so we can write this

as 3 b + a into a + b into b + 3

a upon 32 and this is AB BQ will be equal to 5x8 now we know that this 3 A + B A + B

A + 3 B upon 32 it is equal to 15x2 so we'll replace this value here with 15x 2 so we can write 15x

2 AB CU will be equal to 5 by8 now this is 9 this is three so we'll get AB

CU equal 27 or the value of ab is 3 now it say A and B are positive integers so if a and b are positive integers the

only solutions we'll get is either a is 1 and B is 3 or a is 3

and b is one now the question is if n is root of this

equation and if n HMS are inserted between a and C show that difference between the first first and last mean is

equal to a c into a minus C and the question it says n is root of the

equation x² into 1 - A C - x a s + c²

- 1 + AC = 0 so that means n is going to satisfy this equation so we'll get n² into

1 - a - n a s + c² - 1 + a C = 0 so that's our first

equation now it says n hm are inserted between a and C so in between a and C we have

inserted in HMS or we can say between 1 Upon A and 1 upon C we have inserted

n a now this is suppose H1 H2 and hn and corresponding AP is A1 A2 and a n now for this corresponding

AP its common difference will be 1 upon c - 1 Upon A upon n + 1 now this A1 it will be 1 by

a + D which is 1x a + 1 by c - 1X a upon n + 1 which is n Upon A + 1 upon C upon n + 1 so

that's your A1 and a n will will be 1 upon C minus D

and there'll be 1 upon C and then - D will be - 1 upon c - 1 Upon A upon n + 1 and there'll be n upon C + 1 Upon A upon

n + 1 now their corresponding harmonic means will be H1 which is n + 1

upon N C + a into AC and hn will

be n + 1 AC upon N A + C now it says we need to show that difference between first and last

mean is equal to a c into a minus C we need to find H1 - hn so we can write H1 - hn

is now this n +1 AC it be common and then we'll have 1 upon N C + a - 1 upon N A +

C now if we take LCM we can WR n +1 AC and here it will be na a + C

minus n c minus a upon and this is n²

a + n a s + c² + a

c now we can write this is n + 1 a here we'll take terms of n together so there'll be

N A minus C and plus C minus a and it will be n² a +

n a s + c² now what we'll do is we'll take a minus C common so we'll

get n + 1 a c into a minus C and here will be this nus1

upon n² a + n a s + c² + a now this is A + B into a minus B so we can write this is n² -1 a

c into a minus C Upon n² A C + n a² + c² + a now now we need to prove that this value it is

equal to AC into a minus C which simply means that value of this expression it must be

one and here there is this correction it is AC so it should be this AC so it should be AC it should also be AC and

this is also AC now we know that n is the root of this given equation so you can write

n² into 1 - A C - n a s + c² and then - 1 + a = 0 now we'll take this n² and 1 here and we'll

take everything else on the right hand side so we can write n² - 1 and that will be equal to n² a +

n a s + c² + AC and if we divide it we'll get this result n² - 1 Upon n² A C + n a² +

c² + a c = 1 and if you put it here we'll get this is simply a a minus C which is what we need to prove in this

question now here the questions we are given two series and we have to find the sum of this third series which is 1 +

3ab + 5 a² up to infinite now this x is 1 + 3 a + 6 a s + 10 a CU up to infinite now

what we'll do is we'll write this series again and we'll multiply every term with a and we'll displace this term by one so

we'll get this a and this is 3 a² next one will be 6 a cub up to infinite and we'll subtract second from first we'll

get 1 - a x and will be equal to 1 + 2 a + 3 a² + 4 a CU up to infinite now this is

an AGP now we write the series again and again multiplying it with a and

displacing by one term so we'll get a into 1 - a x and here will be this a and this is 2 a

s + 3 a CU up to infinite now we subtract second from first we'll get 1 - a² into X and here will be this 1 +

a + a² + a q up to infinite which is an infinite GP which is a upon 1 - R so from here we can write X is 1

upon 1 - a Q and from here we can write 1 - A is

x^ - 1x3 or a is 1 - x^ - 1X 3 and that's our first result now this second series

is y = 1 + 4 b + 10 b² + 20 b c up to infinite now again we'll

multiply it with this common ratio B and display repl it by 1 so we'll get b y will be equal to B + 4 b² + 10 BQ up to

infinite so it'll be this 1 - b y and this is 1 + 3 b + 6 b² + 10 BQ up to infinite now this

series it is same as this first series 1 + 3 a + 6 a s + 10 a cub and we know that answer to this series is 1 upon 1 -

A CU so answer to the series will be 1 upon 1 - BQ or Y is

simply 1 upon 1 - b^ 4 from here we can write B is 1 - y^ - 1 by 4 and that's our

second result so we have the value of a and b now we have to find sum of the series

which is 1 + 3 a

+ 5 a² up to infinite in terms of X and Y so again we let the same series and

multiply it with common difference ab and displacing it by one so it'll be this a and this is

3 a² to infinite now we subtract second from first we'll get 1 - AB into s and there'll be 1 + 2

a + 2 a² up to infinite and this is nothing but an infinite GP so we can write this

as 1 - A into s and this is 1 + and this is 2 AB upon 1 -

A which is 1 + a upon 1 - A we can write this S as 1 + a

upon 1 - a² where a is 1 - x^ - 1 by 3 and b is 1 - y^ - 1

by 4 if we put the value of a and b we'll get the value of s in terms of X and Y which is what we need to find in

this question now here it says the real numbers X1 X2 and X3 satisfying the [Music]

equation x cubus x² squ + beta x + GMA = 0 there are in AP then

find the intervals in which beta and gamma lies now suppose these three Roots X1 X2 and X3 which are in AP they

are a minus D A and A + D now if we write S1 which is sum of roots we can write

3 A = 1 or the value of a is 1x 3 now we write S2 which is sum of Roots taken two at a time so it'll be this a

into a minus D plus a into a + D+ Aus D into a + D and that'll be equal to Beta now here a will cancel so we'll

get this as 2 * a² + a² - D ² it is equal to Beta or we can write d s

is 3 a² - beta now a is 1 by 3 so we'll get this as 1X 3 minus beta now D ² it should be

greater than or equal to Z so from here we'll get this condition that value of beta must be less than or equal to

1x3 now we need to find a range of gamma now for gamma we write S3 now S3 is product of roots so this is

a a minus D into a + D and it will be equal to minus

GMA now this is a into a sare - D sare and this is minus gamma now a is 1 by 3 so it'll be this

1X 3 and this is 1x 9 - D ² and this is minus gamma we can

write d s is 1x 9 + 3 GMA now again D ² it should be

greater than or equal to Z so from here we'll get value of gamma it should be greater than or equal to

minus 1 by 27 so we have range of beta and we also have range of gamma and that is the

answer to this question now the question is for an odd integer n we have to find the value

of n minus N - 1 CU and goes all the way up to -1 to the power n -1 into 1 CU and suppose this sum is s

now we know that this n is an odd integer and if n is odd then -1 the power nus1 it be even so it be this +

one so we can also write this as 1 Q - 2 Q + 3 Q - 4 q I'll go all the way up to minus

n -1 cu+ NB now what we'll do is now what we'll do is we'll

add 2 Cube and we'll also subtract 2 cube in the same way we'll add 4 Cube and we'll subtract 4 CU and we'll do it

for all the negative numbers plus N - 1 CU and then minus n -1 CU and what we'll do is we'll take

all these positive terms together so this s will become 1 CU + 2 CU + 3 CU up to n

CU and we'll take all the negative terms together so here we have - 2 cub and - 2 Cub so there'll be - 2

into 2 Q + 4 q up to n minus 1 Q now we can write this is 1 Q +

2 Q up to n CU and here we'll take 2 Cube common so

there'll be 16 and here we'll have 1 Cub + 2 Cub up to N - 1X 2 Cub now since n is odd n - 1

upon 2 it is an integer so there is no problem using this value now this is 1 Cub 2 Cub up to n Cub so it'll be this n

n + 1 by 2 squ and this is - 16 and this is 1 Cub 2 Cub up to N - 1 by 2 Cub so there'll

be n into n + 1 by 2 whole

Square now 2 square is 4 so this 4 4 16 16 will cancel and we can take n + 1 upon 2 whole Square Commons we'll

take n + 1 upon 2 whole s common and what remains is this n² minus N - 1 s now this is A + B into a

minus B so it'll be this 2 N - 1 so it'll be this n + 1 by 2 s

into 2 nus1 and that is the answer to this question now the question it says if a BC are first three terms of a

geometric Series so we are given a geometric progression or geometric series which is a b and c now rather

than taking them as a b and c we'll take them as a a r and a² now it says harmonic mean of A and B is 12 so their

harmonic mean is 12 and that of B and C is 36 now harmonic mean of a and AR R is 2 a into a r Upon

A 1 + r and that'll be equal to 12 now here a it will cancel so we can

write a r = 6 * 1 + r and that's our first equation and harmonic mean of a r and a r Square will be 2

* a r into a r squ upon a r + a r squ so we'll take a r common and it will be this 1 +

r and that will be 36 now this a r it will cancel and we know that a r upon 1 + r it is 6 so we can write 2 *

R into 6 it is equal to 36 so value of R is 3 and if we put RS3 we'll

get 3 a and this is 6 into 4 or value of a is 8 so we have a and r now we need to write

first five terms so it will be 8 24

72 216 and 648 and that is the answer to this

question now here the question is for an AP nth term is 1 byn

and nth term is 1 by m you would find its MN term it's very simple question now nth term is a

+ m - 1 into D and it is 1 by m and nth term is a + N - 1 into D it is 1 by m and we

subtract second from first we get M - n into D will be equal 1 by

nus 1 by m which is M - n upon MN and since M and N they are unequal we'll get DS 1 by MN now if we put DS 1 by MN we

get a plus M -1 upon MN and it is is equal to 1 by m from

here we get a also as 1 by MN and once we have a n d we can write its MN term its MN term will be a