The fifth term of a G . P . is 81 and the second term is 24 , find the series.

Let us assume that a is the first term and r is the common ratio of the given G.P. then for a G . P . , n th term = ar n - 1 (formula) As given in the question, ∴ 81 = a r 4 . . . ( i ) and 24 = a r . . . ( ii ) Dividing ( i ) from ( ii ) , we get a r 4 a r = 81 24 Cancelling a r from numerator and denominator (as a r ≠ 0 ), we get r 3 = 27 8 = 3 2 3 ∴ r = 3 2 Substituting this value of r in ( ii ) , we get a · 3 2 = 24 or a = 16 Hence the required sequence is 16 , 24 , 36 , … . . .

If S p denotes the sum of the series 1 + r p + r 2 p + … … ∞ and s p be the sum of the series 1 - r p + r 2 p - r 3 p + … … ∞ , then prove that S p + s p = 2 S 2 p .

Here in the question, given for first sequence is a = 1 and common ratio r = r p Let us replace p by 2 p in the first equation, we get S 2 p = 1 + r 2 p + r 4 p + . . . . ∞ Let us add S p and s p to get S p + s p = 2 + 2 r 2 p + 2 r 4 p + . . . . ∞ = 2 1 + r 2 p + r 4 p + . . . . . ∞ = 2 S 2 p ∴ S p + s p = 2 S 2 p

E M B I B E

Higher Algebra>Chapter 5 - Geometrical Progression>Examples. V.a.

H S Hall and S R Knight Solutions for Chapter: Geometrical Progression, Exercise 1: Examples. V.a.

Author:H S Hall & S R Knight

H S Hall Mathematics Solutions for Exercise - H S Hall and S R Knight Solutions for Chapter: Geometrical Progression, Exercise 1: Examples. V.a.

Attempt the practice questions on Chapter 5: Geometrical Progression, Exercise 1: Examples. V.a. with hints and solutions to strengthen your understanding. Higher Algebra solutions are prepared by Experienced Embibe Experts.

Questions from H S Hall and S R Knight Solutions for Chapter: Geometrical Progression, Exercise 1: Examples. V.a. with Hints & Solutions

EASY

JEE Main

IMPORTANT

Higher Algebra>Chapter 5 - Geometrical Progression>Examples. V.a.>Q 21

The fifth term of a $G . P .$ is $a \cdot \frac{3}{2} = 24$ and the second term is $24$ , find the series.

EASY

JEE Main

IMPORTANT

Higher Algebra>Chapter 5 - Geometrical Progression>Examples. V.a.>Q 22

Find the first term of the $G . P .$ having common ratio $3$ , last term $486$ and sum of the terms equal to $728$ .

EASY

JEE Main

IMPORTANT

Higher Algebra>Chapter 5 - Geometrical Progression>Examples. V.a.>Q 23

In a $G . P .$ , the first term is $7$ , the last term $448$ , and the sum $889$ , find the common ratio.

EASY

JEE Main

IMPORTANT

Higher Algebra>Chapter 5 - Geometrical Progression>Examples. V.a.>Q 24

The sum of three consecutive numbers in $G . P .$ is $38$ , and their product is $1728$ , find the terms.

EASY

JEE Main

IMPORTANT

Higher Algebra>Chapter 5 - Geometrical Progression>Examples. V.a.>Q 25

The continued product of three numbers in $G . P .$ is $216$ , and the sum of the products of them in pair is $156$ , find the numbers.

MEDIUM

JEE Main

IMPORTANT

Higher Algebra>Chapter 5 - Geometrical Progression>Examples. V.a.>Q 26

If $S_{p}$ denotes the sum of the series $1 + r^{p} + r^{2 p} + \dots \dots \infty$ and $s_{p}$ be the sum of the series $1 - r^{p} + r^{2 p} - r^{3 p} + \dots \dots \infty$ , then prove that $S_{p} + s_{p} = 2 S_{2 p}$ .