13 - Geometric Sequence and Series (Algebra)

In this post, you can check the solutions to the exercises of Geometric Sequence and Series of Algebra chapter of Readmore Optional Mathematics by DR Simkhada. It consists of complete answers to all odd questions mentioned in the textbook.

Before starting, make sure to check the following note of Geometric Sequence and Series.

Things to remember:

  1. The first term of the sequence is denoted by $t_1$ or a.
  2. The common ratio of the sequence is denoted by r.
  3. Total number of terms in the sequence is denoted by n.
  4. The general formula for the $n^{th}$ term of a GS is $t_n = ar^{n-1}$.


Exercise 1.3

1. The common ratio and first term of the geometric sequence are given below. Find the first three terms of the sequence.

a) First term (a) = 3 and common ratio (r) = 2
Solution:

In a geometric sequence,
$t_n = ar^{n-1}$
$or, t_n = 3×2^{n-1}$

When n = 1,
$t_1 = a = 3$

When n = 2,
$t_2 = 3×2^{2-1}$
$or, t_2 = 3×2$
$\therefore t_2 = 6$

When n = 3,
$t_3 = 3×2^{3-1}$
$or, t_3 = 3×2^2$
$or, t_3 = 3×4$
$\therefore t_3 = 12$
Hence, the required first three terms of the sequence are 3,6 and 12.


c) a = -8 and r = 4
Solution:
$t_n = ar^{n-1}$
$t_1 = a = -8$
$t_2 = ar = (-8)×4 = -32$
$t_3 = ar^2 = (-8)×4^2 = -128$

Hence, the required first three terms of the sequence are -8, -32 and -128.


2. Calculate the common ratio (r) and 8th term in the following sequence.

a) 32,16,8,4,...
Solution:
Here,
Geometric sequence is 32,16,8,4,..
$a = 32$
$r = \frac{t_2}{a} = \frac{16}{32} = \frac{1}{2}$

We know,
$t_n = ar^{n-1}$
So,
$t_8 = ar^7$
$= 32 × \left ( \dfrac{1}{2} \right )^7$

$= 32 × \dfrac{1}{128}$
$= \dfrac{1}{4}$
Hence, the required common ratio and eighth term of the sequence are 1/2 and 1/4, respectively.


3. Calculate the first term of the geometric sequence whose:

a) r = 3 and t5 = 162
Solution:
Here,
$r = 3$
$t_5 = 162$
To find: a = ?
We have,
$t_n = ar^{n-1}$
$or, t_5 = ar^4$
$or, 162 = a×3^4$
$or, 162 = 81a$
$\therefore a = 2$
Hence, the required first term of the sequence is 2.


4. Calculate the positive common ratio of the geometric sequence whose:

a) a = 16 and t4 = 128
Solution:
Here,
$a = 16$
$t_4 = 128$
To find: r = ? [r is positive]

We have,
$t_n = a r^{n-1}$
$or, t_4 = ar^3$
$or, 128 = 16 × r^3$
$or, r^3 = 8$
$\therefore r = 2$
Hence, the required positive common ratio of the given sequence is 2.


5. How many terms are there in the following series?

a) 4+8+16+...+1024
Solution:
Series:4+8+16+...+1024
Here,
$a = 4$
$r = \frac{t_2}{a} = \frac{8}{4} = 2$
$t_n = 1024$
We know,
$t_n = ar^{n-1}$
$or, 1024 = 4×2^{n-1}$
$or, 256 = 2^{n-1}$
$or, 2^8 = 2^{n-1}$
$or, 8 = n-1$
$\therefore n = 8+1 = 9$
Hence, there are 9 terms in the given series.


6. Calculate the positive common ratio and first term of the given series whose,

a) t4 = 24, t7 = 192
Solution:
Here,
$t_4 = 24$
$t_7 = 192$
We know,
$t_n = ar^{n-1}$
Using formula,
$t_4 = ar^3$
$or, 24 = ar^3$
$or, a = \dfrac{24}{r^3}$ - (i)

And,
$t_7 = ar^6$
$or, 192 = ar^6$
Put value of a from equation (i), we get,

$or, 192 = \dfrac{24}{r^3} × r^6$

$or, \dfrac{192}{24} = r^{6-3}$

$or, 8 = r^3$
$\therefore r = 2$
Also,
Put value of r = 2 in equation (i), we get,
$or, a = \dfrac{24}{2^3}$
$or, a = \dfrac{24}{8}$
$\therefore a = 3$
Hence, the required positive common ratio and first term of the given series are 2 and 3, respectively.


7 a) 6th and 8th terms of a GS having positive 'r' are 64 and 256 respectively. Find t11.

Solution:
Here,
$t_6 = 64$
$t_8 = 256$

We know,
$t_n = ar^{n-1}$

Using formula,
$t_6 = ar^5$

$or, 64 = ar^5$

$or, a = \dfrac{64}{r^5}$ - (i)

Also,
$t_8 = ar^7$

$or, 256 = ar^7$

Put value of a from equation (i), we get,

$or, 256 = \dfrac{64}{r^5} × r^7$

$or, \dfrac{256}{64} = r^{7-5}$

$or, 4 = r^2$

$\therefore r = 2$ [since, r is positive]

Put value of r in equation (i), we get,
$or, a = \dfrac{64}{2^5}$

$\therefore a = 2$

So,
$t_11 = ar^{10} = 2×2^{10} = 2048$

Hence, the required value of the 11th term is 2048.

About Readmore Optional Mathematics Book 10

Author: D. R. Simkhada
Editor: I. R. Simkhada

Readmore Publishers and Distributors
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Geometric Sequence and Series - Class 10 Solved Exercises | Readmore Optional Mathematics is a collection of the solutions related to exercises of Geometric Sequence and Series chapter from Algebra Unit for Nepal's Secondary Education Examination (SEE) appearing students.

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