# Sequence and Series Problems

## 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 ai. 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 SN = a1 + a2 + a3 + a4 +……. + an.

## 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 the one with finite terms whereas an infinite sequence is with never ending terms or infinite in count.
• Arithmetic Sequence: A sequence where the sum or difference between any two consecutive terms is a constant, termed as 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:

• The nth term of the arithmetic sequence or arithmetic progression (A.P) is given by an = a + (n – 1) d
• The arithmetic mean [A.M] between a and b is A.M = [a + b] / 2
• The nth term an of the geometric sequence or geometric progression [G.P] is an = a * rn–1
• The geometric mean between a and b is G.M= $\pm \sqrt{ab}$
• The nth term an of the harmonic progression is an= 1 / [a + (n – 1) d]
• The harmonic mean between a and b is H.M = 2ab / [a + b]
• Series of A.P : Sn = [n / 2] (a + l), Sn = (n / 2) [2a + (n – 1) d], where Sn is sum to n terms of A.P.
• The sum of ‘n’ A.M between a and b is A.M = [n (a + b)] / 2
• Series of G.P: Sn = [a (1 – rn)] / [1 – r] ; where Sn is the sum to n terms of G.P.
• The sum ‘S’ of infinite geometric series is S = a / 1–r.

## Solved Examples On Sequence and Series

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

11the term is given by ar10 = 6×210 = 6144

Hence the required answer is 6144.

Example 2: If Sn denotes the sum of first n terms of an A.P. and [S3n − Sn−1] / [S2n −S2n−1] = 31, then the value of n is __________.

Solution:

S3n = [3n / 2] [2a + (3n − 1) d]

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

⇒ S3n − Sn−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]

S2n − S2n−1 = T2n = a +(2n − 1) d

⇒ S3n − Sn − S2n −S2n−1 = (2n + 1)

Given, ⇒ [S3n − Sn−1] / [S2n − S2n−1] = 31

⇒ n = 15

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

Solution:

If tr is the rth term of the A.P., then

tr = Sr − Sr−1 = cr (r − 1) − c (r − 1) (r − 2) = c (r − 1) (r − r + 2) = 2c (r − 1)

We have, t12 + t22+……..+tn2 = 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:

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

100th term 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; hence 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 ex = 1 + x / 1! + x2 / 2! + x3 / 3! + x4 / 4! + …….. ∞ ——— (i)

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

On adding equations (i) and (ii),

we get ex + e−x = 2x + 2x2 / 2! + 2x4 / 4!+…∞

∴ [ex + 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 A.P., p, q, and r are in H.P., and ap, bq, and cr are in G.P., then [p / r] + [r / p] is equal to ______________.

Solution:

p / r + r / p = [p2 + r2] / [pr]

= (p + r)2 − ([2pr] / [pr])

= $\frac{\frac{4p^2r^2}{q^2}-2pr}{pr}$

= [(4p2r2 / q2) − 2pr] / [pr]

[p, q, r are in H.P. 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 A.P. where the distinct numbers x, y, z are in G.P., then the common ratio of the G.P. is ___________.

Solution:

x, y, and z are in G.P.

Hence, Y= xr, zy = xr, x = xr2

Also, x, 2y, and 3z are in A.P.

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 $ar,a{{r}^{2}},a{{r}^{3}},…a{{r}^{10}}$. If their sum is 18 and the sum the reciprocals is 6, then the product of these ten numbers is

A) 81

B) 243

C) 343

D) 324

Solution:

Given $\frac{ar({{r}^{10}}-1)}{r-1}=18 \rightarrow (1)$

Also,

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

From (1) and (2)

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

Example 9: If ${{S}_{n}}$ denotes the sum of first n terms of an A.P. whose first term is a and ${{S}_{nx}}/{{S}_{x}}$ is independent of x, then ${{S}_{p}}=$

$A) {{p}^{3}}\\ B) {{p}^{2}}a\\ C) p{{a}^{2}}\\ D) {{a}^{3}}$

Solution:

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

For $\frac{{{S}_{nx}}}{{{S}_{x}}}$ to be independent of x,

2a – d = 0 or

2a = d

Now,

${{S}_{p}}=\frac{p}{2}[2a+(p-1)d]={{p}^{2}}a$

Example 10: If ${{a}_{1}},{{a}_{2}},{{a}_{3}}….{{a}_{n}}$ are in H.P. and $f(k)=\left( \sum\limits_{r=1}^{n}{{{a}_{r}}} \right)-{{a}_{k}}$ then $\frac{{{a}_{1}}}{f(1)},\frac{{{a}_{2}}}{f(2)},\frac{{{a}_{3}}}{f(3)},…\frac{{{a}_{n}}}{f(n)}$ are in

A) A.P

B) G.P

C) H.P

D) none of these

Solution:

$f(k)+{{a}_{k}}=\sum\limits_{r=1}^{n}{{{a}_{r}}}=\lambda \,(\text \ say)\\ f(k)=\lambda -{{a}_{k}}\\ \Rightarrow \frac{{{f}_{(k)}}}{{{a}_{k}}}=\frac{\lambda }{{{a}_{k}}}-1\\ \frac{f(1)}{{{a}_{1}}},\frac{f(2)}{{{a}_{2}}},…,\frac{f(n)}{{{a}_{n}}}$ are in A.P.

So $\frac{{{a}_{1}}}{f(1)},\frac{{{a}_{2}}}{f(2)},…\frac{{{a}_{n}}}{f(n)}$ are in H.P.

Example 11: If ${{x}_{1}},{{x}_{2}},…..{{x}_{20}}$ are in H.P. and ${{x}_{1}},2,{{x}_{20}}$ are in G.P., then $\sum\limits_{r=1}^{19}{{{x}_{r}}{{x}_{r+1}}}=$

Solution:

Clearly

$\frac{1}{{{x}_{1}}},\frac{1}{{{x}_{2}}},…\frac{1}{{{x}_{20}}}$ will be in A.P. Hence,

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

Now,

$\frac{1}{{{x}_{20}}}=\frac{1}{{{x}_{1}}}+19\lambda\\ \frac{{{x}_{1}}-{{x}_{20}}}{{{x}_{1}}{{x}_{20}}}=19\lambda \\ \Rightarrow \sum\limits_{r=1}^{19}{{{x}_{r}}{{x}_{r+1}}=19{{x}_{1}}{{x}_{20}}=19\times 4=76}$

[Because ${{x}_{1}},2,{{x}_{20}}$ are in G.P., then ${{x}_{1}}{{x}_{20}}=4)$.