The first, third, and seventh terms of a non-constant arithmetic progression (AP) form the… — JAMB UTME

The first, third, and seventh terms of a non-constant arithmetic progression (AP) form the first three consecutive terms of a geometric progression (GP). If the first term of the AP is $4$ , what is the sum of the first $4$ terms of the geometric progression?

60Answer
B
28
C
124
D
92

Answer

The sum of the first

4

terms of the geometric progression is

60

The first three terms of the GP are

T_1 = 4

T_3 = 4 + 2d

, and

T_7 = 4 + 6d

. Equating

(4 + 2d)^2 = 4(4 + 6d)

yields

4d^2 - 8d = 0

, giving

d = 2

. The first four terms of the GP are

4, 8, 16, 32

, which sum to

4 + 8 + 16 + 32 = 60

Step-by-Step Solution

Express the terms of the arithmetic progression in terms of first term

a

and common difference

d

First term

T_1 = 4

, third term

T_3 = 4 + 2d

, seventh term

T_7 = 4 + 6d

The

n^{\text{th}}

term of an AP is defined as

T_n = a + (n-1)d

Set up the geometric progression condition

(T_3)^2 = T_1 \cdot T_7

to solve for

d

(4 + 2d)^2 = 4(4 + 6d) \implies 16 + 16d + 4d^2 = 16 + 24d \implies 4d^2 - 8d = 0 \implies d = 2

(since the AP is non-constant,

d \neq 0

Three terms

x, y, z

form a GP if and only if

y^2 = xz

Determine the terms and common ratio

r

of the GP.

First term

G_1 = 4

, second term

G_2 = 4 + 2(2) = 8

. Thus,

r = \frac{8}{4} = 2

The common ratio

r

is the quotient of consecutive terms of the GP.

Calculate the sum of the first

4

terms of the GP using

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

S_4 = \frac{4(2^4 - 1)}{2 - 1} = \frac{4(16 - 1)}{1} = 60

Formula for the sum of the first

n

terms of a geometric progression.

Key Concept

Arithmetic and Geometric Progression Inter-relationships

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