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In a geometric progression, the sum of the first 5 term is 31/3 and the sum of t...

Aug 5, 2024

In a geometric progression, the sum of the first 5 term is 31/3 and the sum of the next 5 terms is 992/3. Find the first term.

Solution by Steps

step 1

Given the sum of the first 5 terms of a geometric series is

\frac{31}{3}

, we can write the equation as

a \left( \frac{1 - r^5}{1 - r} \right) = \frac{31}{3}

step 2

Given the sum of the next 5 terms of the same geometric series is

\frac{992}{3}

, we can write the equation as

a r^5 \left( \frac{1 - r^5}{1 - r} \right) = \frac{992}{3}

step 3

Divide the second equation by the first equation to eliminate

a

\frac{a r^5 \left( \frac{1 - r^5}{1 - r} \right)}{a \left( \frac{1 - r^5}{1 - r} \right)} = \frac{\frac{992}{3}}{\frac{31}{3}}

This simplifies to:

r^5 = \frac{992}{31}

step 4

Solve for

r

r^5 = \frac{992}{31} \implies r = \left( \frac{992}{31} \right)^{\frac{1}{5}}

step 5

Substitute

r

back into the first equation to solve for

a

a \left( \frac{1 - \left( \frac{992}{31} \right)}{1 - \left( \frac{992}{31} \right)^{\frac{1}{5}}} \right) = \frac{31}{3}

step 6

Simplify and solve for

a

a = \frac{31}{3} \left( \frac{1 - \left( \frac{992}{31} \right)^{\frac{1}{5}}}{1 - \left( \frac{992}{31} \right)} \right)

Answer

The first term

a

\frac{31}{3} \left( \frac{1 - \left( \frac{992}{31} \right)^{\frac{1}{5}}}{1 - \left( \frac{992}{31} \right)} \right)

Key Concept

Geometric Series Sum Formula

Explanation

The sum of the first

n

terms of a geometric series can be found using the formula

S_n = a \left( \frac{1 - r^n}{1 - r} \right)

, where

a

is the first term and

r

is the common ratio. By setting up equations for the sums of different segments of the series, we can solve for

a

and

r