The 3^{\text{rd}}, 6^{\text{th}}, and 11^{\text{th}} terms of a non-constant arithmetic pr… — JAMB UTME

The $3^{\text{rd}}$ , $6^{\text{th}}$ , and $11^{\text{th}}$ 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 $15$ , what is the $4^{\text{th}}$ term of the geometric progression?

Answer: 125

Answer

The 4th term of the geometric progression is 125.

By writing the 3rd, 6th, and 11th terms of the AP as

15+2d

15+5d

, and

15+10d

, we utilize the geometric mean property

(15+5d)^2 = (15+2d)(15+10d)

to find

d=6

. This yields the GP terms

27, 45, 75

, giving a common ratio of

5/3

. Multiplying the third term

75

5/3

gives the 4th GP term as

125

Step-by-Step Solution

Write down the AP term expressions

T_3 = 15 + 2d

T_6 = 15 + 5d

T_{11} = 15 + 10d

The

n^{\text{th}}

term of an AP is given by

T_n = a + (n-1)d

with initial term

a = 15

Apply the geometric progression condition

(15 + 5d)^2 = (15 + 2d)(15 + 10d)

If three terms

A, B, C

are in GP, then

B^2 = A \cdot C

Expand and solve the quadratic equation for the common difference

d

d = 6

Expanding gives

225 + 150d + 25d^2 = 225 + 180d + 20d^2 \implies 5d^2 = 30d \implies d = 6

because

d \neq 0

Find the terms and common ratio of the GP

G_1 = 27

G_2 = 45

G_3 = 75

, and common ratio

r = \frac{5}{3}

Substituting

d = 6

gives the GP terms, and dividing consecutive terms gives

r = \frac{45}{27} = \frac{5}{3}

Calculate the 4th term of the GP

G_4 = 125

Multiplying the 3rd term by the common ratio yields

75 \times \frac{5}{3} = 125

Key Concept

Combining Arithmetic Progression nth-term formulas with Geometric Progression consecutive-term properties

Estimated Time:2m 30s

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