Sequence and Series Solved Problems for IIT JEE

Introduction to Sequence and Series

In Mathematics, a sequence is a group of numbers in an ordered way following a pattern. The numbers are the terms, and the index is the position of the term denoted by a_i. In this article, you will learn sequence and series problems along with the solutions.

A series is the summation of a sequence. It is given by S_N = a₁ + a₂ + a₃ + a₄ +……. + a_n.

Types of Sequence and Series

The following are the types of sequence and series.

In a finite series, a finite number of terms are present, whereas an infinite series consists of infinite terms.

Some common sequences are as follows:

Finite and Infinite Sequence: A finite sequence is one with finite terms, whereas an infinite sequence is with never-ending terms or is infinite in the count.
Arithmetic Sequence: A sequence where the sum or difference between any two consecutive terms is a constant, termed a common difference.
Geometric Sequence: A sequence where every term is obtained by multiplying or dividing a certain number by the preceding term.
Harmonic Sequence: A sequence obtained by the reciprocals of the elements of an arithmetic sequence.
Fibonacci Sequence: A sequence where the sum of the two preceding terms will give the succeeding term.

Important Formulas

The formulae for sequence and series are as follows:

The n^th term of the arithmetic sequence or arithmetic progression (A.P) is given by a_n= a + (n – 1) d
The arithmetic mean [AM] between a and b is AM = [a + b] / 2
The n^th term an of the geometric sequence or geometric progression [GP] is a_n= a * r^n–1
The geometric mean between a and b is
\(\begin{array}{l}GM = \pm \sqrt{ab}\end{array} \)
The n^th term a_n of the harmonic progression is a_n= 1 / [a + (n – 1) d]
The harmonic mean between a and b is HM = 2ab / [a + b]
Series of AP: S_n = [n / 2] (a + l), S_n = (n / 2) [2a + (n – 1) d], where S_n is the sum to n terms of AP.
The sum of ‘n’ AM between a and b is AM = [n (a + b)] / 2
Series of GP: S_n = [a (1 – rⁿ)] / [1 – r], where S_n is the sum to n terms of GP.
The sum ‘S’ of infinite geometric series is S = a / 1–r.

Also, Read:

Solved Examples on Sequence and Series

Example 1: In a GP, the second term is 12, and the sixth term is 192. Find the 11^th term.

Solution:

Given second term, ar = 12 ..(1)

Sixth term, ar⁵ = 192 …(2)

Dividing (2) by (1), we get ar⁵ / ar = 192/12

⇒ r⁴ = 16

⇒ r = 2

Substitute r in (1), we get a×2 = 12

⇒ a = 12/2 = 6

11^theterm is given by ar¹⁰= 6×2¹⁰ = 6144

Hence, the required answer is 6144.

Example 2: If S_n denotes the sum of the first n terms of an AP and [S_3n − S_n−1] / [S_2n −S_2n−1] = 31, then the value of n is __________.

Solution:

S_3n= [3n / 2] [2a + (3n − 1) d]

S_n−1= ([n − 1] / 2) * [2a + (n − 2) d]

⇒ S_3n− S_n−1= [1 / 2] [2a (3n − n + 1)] + [d / 2] [3n (3n − 1) − (n − 1) (n − 2)]

= [1 / 2] [2a (2n + 1) + d (8n²− 2)]

= a (2n + 1) + d (4n² − 1)

= (2n + 1) [a + (2n − 1) d]

S_2n − S_2n−1= T_2n= a +(2n − 1) d

⇒ S_3n− S_n − S_2n −S_2n−1= (2n + 1)

Given, ⇒ [S_3n − S_n−1] / [S_2n − S_2n−1] = 31

⇒ n = 15

Example 3: If the sum of n terms of an AP is (c*n) (n – 1), where c ≠ 0, then the sum of the squares of this term is __________.

Solution:

If t_r is the r^th term of the AP, then

t_r = S_r − S_r−1 = cr (r − 1) − c (r − 1) (r − 2) = c (r − 1) (r − r + 2) = 2c (r − 1)

We have, t₁² + t₂²+……..+t_n²= 4c² (o² + 1² + 2²… + (n − 1)²)

= [4c²] * [(n − 1) n (2n − 1)] / [6]

= [2 / 3] c² n (n − 1) (2n − 1)

Example 4: The largest term common to the sequences 1, 11, 21, 31,…. to 100 terms and 31, 36, 41, 46,………to 100 terms is ___________.

Solution:

100^thterm of 1, 11, 21, 31, …………. is 1+ (100 – 1) 10 = 991.

100^thterm of 31, 36, 41, 46, ………… is 31 + (100 – 1) 5 = 526.

Let the largest common term be 526.

Then, 526 = 31 + (n – 1) 10

Or n = 50.5

But n is an integer, so n = 50.

Hence, the largest common terms are 31 + (n – 1) 10 = 521.

Example 5: Find the sum of the series 1 + 1 / [4 * 2!] + 1 / [16 * 4!] + 1 / [64 * 6!]+……….

Solution:

We know that e^x= 1 + x / 1! + x²/ 2! + x³ / 3! + x⁴ / 4! + …….. ∞ ——— (i)

∴ [e^x + e^−x] / 2 = 1 + x² / 2! + x³ / 3! + x⁴ / 4………..∞ ————-(ii)

On adding equations (i) and (ii),

We get e^x + e^−x= 2x + 2x² / 2! + 2x⁴ / 4!+…∞

∴ [e^x + e^−x] / 2 = 1 + x² / 2! + x⁴ / 4 + x⁶ / 6!+ ……………

Putting x = 1 / 2, we get

e + 1 / 2√ e = 1 + 1 / [4 * 2!] + 1 / [16 * 4!] + 1 / [64 * 6!]+……….

Example 6: If a, b, and c are in AP, p, q, and r are in HP, and ap, bq, and cr are in GP, then [p / r] + [r / p] is equal to ______________.

Solution:

p / r + r / p = [p² + r²] / [pr]

= (p + r)² − ([2pr] / [pr])

\(\begin{array}{l}=\frac{\frac{4p^2r^2}{q^2}-2pr}{pr}\end{array} \)

= [(4p²r² / q²) − 2pr] / [pr]

[p, q, r are in HP. Therefore, q = 2pr / [p + r]]

= [4pr / q²] − [2]

= [4b / ac] − [2]

[ap, bq, cr are in A.P ⇒ b2q2 = ac * pr]

= [(a + c)² / ac] − [2]

[a, b, c are in A.P ⇒ 2b = a + c]

= [a / c] + [c / a].

Example 7: If x, 2y, 3z are in AP, where the distinct numbers x, y, z are in GP, then the common ratio of the GP is ___________.

Solution:

x, y and z are in GP.

Hence, Y= xr, zy = xr, x = xr²

Also, x, 2y, and 3z are in AP.

Hence, 4y = x + 3z ⇒4xr = x + 3xr²

⇒ 3r² − 4r + 1 = 0

⇒ (3r − 1) (r − 1) = 0

⇒ r = 1/3

Example 8: Consider the ten numbers ar, ar², ar³,…, ar¹⁰. If their sum is 18 and the sum of the reciprocals is 6, then the product of these ten numbers is __________.
A) 81

B) 243

C) 343

D) 324

Solution:

Given

\(\begin{array}{l}\frac{ar({{r}^{10}}-1)}{r-1}=18 \rightarrow (1)\end{array} \)

Also,

\(\begin{array}{l}\frac{\frac{1}{ar}\left( 1-\frac{1}{{{r}^{10}}} \right)}{1-\frac{1}{r}}=6 \\ \frac{1}{a{{r}^{11}}}.\frac{({{r}^{10}}-1)r}{r-1}=6\\ \frac{1}{{{a}^{2}}{{r}^{11}}}.\frac{ar({{r}^{10}}-1)}{r-1}=6 \rightarrow (2)\end{array} \)

From (1) and (2),

\(\begin{array}{l}\frac{1}{{{a}^{2}}{{r}^{11}}}.\times 18=6\\ {{a}^{2}}{{r}^{11}}=3\\ \text Now \ P={{a}^{10}}{{r}^{11}}={{({{a}^{2}}{{r}^{11}})}^{5}}={{3}^{5}}=243\end{array} \)

Example 9: If S_n denotes the sum of the first n terms of an AP, whose first term is a and S_nx/S_x is independent of x, then S_p =

\(\begin{array}{l}A) {{p}^{3}}\\ B) {{p}^{2}}a\\ C) p{{a}^{2}}\\ D) {{a}^{3}}\end{array} \)

Solution:

\(\begin{array}{l}\frac{{{S}_{nx}}}{{{S}_{x}}}=\frac{\frac{nx}{2}[2a+(nx-1)d]}{\frac{x}{2}[2a+(x-1)d]}=\frac{n[(2a-d)+nxd]}{(2a-d)+xd}\end{array} \)

For S_nx/S_x to be independent of x,

2a – d = 0 or

2a = d

Now,

\(\begin{array}{l}{{S}_{p}}=\frac{p}{2}[2a+(p-1)d]={{p}^{2}}a\end{array} \)

Example 10: If a₁, a₂, a₃, …., a_n are in H.P. and

\(\begin{array}{l}f(k)=\left( \sum\limits_{r=1}^{n}{{{a}_{r}}} \right)-{{a}_{k}}\ \text{then}\ \frac{{{a}_{1}}}{f(1)},\frac{{{a}_{2}}}{f(2)},\frac{{{a}_{3}}}{f(3)},…\frac{{{a}_{n}}}{f(n)}\ \text{are in}\end{array} \)

A) AP

B) GP

C) HP

D) none of these

Solution:

\(\begin{array}{l}f(k)+{{a}_{k}}=\sum\limits_{r=1}^{n}{{{a}_{r}}}=\lambda \,(\text \ say)\\ f(k)=\lambda -{{a}_{k}}\\ \Rightarrow \frac{{{f}_{(k)}}}{{{a}_{k}}}=\frac{\lambda }{{{a}_{k}}}-1\\ \frac{f(1)}{{{a}_{1}}},\frac{f(2)}{{{a}_{2}}},…,\frac{f(n)}{{{a}_{n}}}\end{array} \)

are in A.P.

\(\begin{array}{l}\frac{{{a}_{1}}}{f(1)},\frac{{{a}_{2}}}{f(2)},…\frac{{{a}_{n}}}{f(n)}\end{array} \)

are in H.P.

Example 11: If x₁, x₂,…., x₂₀ are in HP and x₁, 2, x₂₀ are in GP, then

\(\begin{array}{l}\sum\limits_{r=1}^{19}{{{x}_{r}}{{x}_{r+1}}}=\end{array} \)

Solution:

Clearly

\(\begin{array}{l}\frac{1}{{{x}_{1}}},\frac{1}{{{x}_{2}}},…\frac{1}{{{x}_{20}}}\end{array} \)

will be in A.P. Hence,

\(\begin{array}{l}\frac{1}{{{x}_{2}}}-\frac{1}{{{x}_{1}}}=\frac{1}{{{x}_{3}}}-\frac{1}{{{x}_{2}}}=…=\frac{1}{{{x}_{r+1}}}-\frac{1}{{{x}_{r}}}=…\lambda (\text{say})\\ \Rightarrow \frac{{{x}_{r}}-{{x}_{r+1}}}{{{x}_{1}}{{x}_{r+1}}}=\lambda \\ {{x}_{r}}{{x}_{r+1}}=-\frac{1}{\lambda }({{x}_{r+1}}-{{x}_{r}})\\ \Rightarrow \sum\limits_{r=1}^{19}{{{x}_{r}}{{x}_{r+1}}=-\frac{1}{\lambda }\sum\limits_{r=1}^{19}{({{x}_{r+1}}-{{x}_{r}})}}\\ =-\frac{1}{\lambda }({{x}_{20}}-{{x}_{1}})\end{array} \)

Now,

\(\begin{array}{l}\frac{1}{{{x}_{20}}}=\frac{1}{{{x}_{1}}}+19\lambda\\ \frac{{{x}_{1}}-{{x}_{20}}}{{{x}_{1}}{{x}_{20}}}=19\lambda \\ \Rightarrow \sum\limits_{r=1}^{19}{{{x}_{r}}{{x}_{r+1}}=19{{x}_{1}}{{x}_{20}}=19\times 4=76}\end{array} \)

[Because x₁, 2, x₂₀are in GP, then x₁x₂₀= 4]

Sequence and Series Problems

Introduction to Sequence and Series

Types of Sequence and Series

Important Formulas

Also, Read:

Top Questions of Sequence and Series for JEE Advanced

Solved Examples on Sequence and Series

Sequence and Series – Revision Video

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