In a geometric progression of positive terms, the sum of the first two terms is 12 and the… — JAMB UTME

In a geometric progression of positive terms, the sum of the first two terms is $12$ and the sum of the third and fourth terms is $48$ . What is the $6^{\text{th}}$ term of the progression?

Answer: 128

Answer

The 6th term of the geometric progression is 128.

Dividing

ar^2(1+r) = 48

a(1+r) = 12

yields

r^2 = 4

, so

r = 2

for positive terms. Substituting

r = 2

into

a(1+r) = 12

gives

a = 4

. Using

T_n = a r^{n-1}

for

n=6

, we get

T_6 = 4 \times 2^5 = 128

Step-by-Step Solution

Set up algebraic equations for the given sums using first term

a

and common ratio

r

a(1+r) = 12

and

ar^2(1+r) = 48

The terms of a geometric progression are given by

T_n = a r^{n-1}

Divide the equation for the third and fourth terms by the equation for the first and second terms.

r^2 = 4 \implies r = 2

Dividing eliminates

a

and

(1+r)

, giving

r^2 = 4

. Since terms are positive,

r > 0

Substitute

r = 2

into

a(1+r) = 12

to solve for

a

a = 4

3a = 12

leads directly to

a = 4

Evaluate the 6th term using the formula

T_6 = a r^{5}

T_6 = 4 \times 2^5 = 128

Applying the general term formula

T_n = a r^{n-1}

with

n=6

Key Concept

Geometric Progression term relations and finding the common ratio from consecutive term pairs

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