The 2^{\text{nd}}, 5^{\text{th}}, and 14^{\text{th}} terms of a non-constant arithmetic pr… — JAMB UTME

The $2^{\text{nd}}$ , $5^{\text{th}}$ , and $14^{\text{th}}$ terms of a non-constant arithmetic progression (A.P.) form three consecutive terms of a geometric progression (G.P.). If the first term of the A.P. is $3$ , what is the sum of the first $6$ terms of the A.P.?

108Answer
B
126
C
99
D
90

Answer

108

With first term

a=3

, the terms

T_2 = 3+d

T_5 = 3+4d

, and

T_{14} = 3+13d

form a geometric progression. Therefore,

(3+4d)^2 = (3+d)(3+13d)

. Expanding gives

9 + 24d + 16d^2 = 9 + 42d + 13d^2

, which simplifies to

3d^2 - 18d = 0

. Since the sequence is non-constant (

d \neq 0

d = 6

. Using the A.P. sum formula

S_n = \frac{n}{2}[2a + (n-1)d]

S_6 = 3[2(3) + 5(6)] = 3[6 + 30] = 108

Step-by-Step Solution

Express the

2^{\text{nd}}

5^{\text{th}}

, and

14^{\text{th}}

terms of the A.P. in terms of the first term

a=3

and common difference

d

T_2 = 3 + d

T_5 = 3 + 4d

, and

T_{14} = 3 + 13d

The

n^{\text{th}}

term of an A.P. is given by

T_n = a + (n-1)d

Set up the condition for consecutive terms of a G.P. and solve for

d

(3 + 4d)^2 = (3 + d)(3 + 13d) \implies 9 + 24d + 16d^2 = 9 + 42d + 13d^2 \implies 3d^2 - 18d = 0 \implies d = 6

For consecutive terms in a G.P., the middle term squared equals the product of the adjacent terms (

T_5^2 = T_2 \times T_{14}

Calculate the sum of the first

6

terms of the A.P. using

a=3

and

d=6

S_6 = \frac{6}{2}[2(3) + (6-1)(6)] = 3[6 + 30] = 3 \times 36 = 108

The sum formula for an A.P. is

S_n = \frac{n}{2}[2a + (n-1)d]

Key Concept

Simultaneous conditions connecting A.P. and G.P. terms combined with sequence sum formulas.

Estimated Time:2m 0s

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