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JAMB Mathematics 2011 Paper

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Question : 19

Total: 46

The seconds term of a geometric series is 4 while the fourth term is 16. Find the sum of the first five terms

60
62
54
64

Solution:

Concept:

The

n

th term of a geometric progression is

T_n = ar^{n-1}

, and the sum of the first

n

terms is

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

.

Explanation:

Given

T_2 = 4

, so

ar^{2-1} = ar = 4

.

Given

T_4 = 16

, so

ar^{4-1} = ar^3 = 16

.

Divide

T_4

by

T_2

:

\frac{ar^3}{ar} = \frac{16}{4}

.

This gives

r^2 = 4

, hence

r = 2

.

Since

ar = 4

, then

a = \frac{4}{r} = \frac{4}{2} = 2

.

Use the sum formula:

S_5 = \frac{2(2^5 - 1)}{2 - 1}

.

Simplify:

S_5 = \frac{2(32 - 1)}{1} = 2(31) = 62

.

Answer:

B. 62

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