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 (AP) form the first three consecutive terms of a geometric progression (GP). If the sum of the first $6$ terms of the AP is $72$ , calculate the common difference of the AP.

Cevap: 4

Cevap

The common difference of the arithmetic progression is

4

Equating the square of the middle GP term

(a+4d)^2

to the product of the outer terms

(a+d)(a+13d)

yields

3d^2 = 6ad

, which simplifies to

d = 2a

. Substituting

2a = d

into the AP sum formula

S_6 = 3(2a+5d) = 72

gives

3(6d) = 72 \implies 18d = 72

, so

d = 4

Adım Adım Çözüm

Express the

2^{\text{nd}}

5^{\text{th}}

, and

14^{\text{th}}

terms of the AP algebraically

T_2 = a + d

T_5 = a + 4d

T_{14} = a + 13d

The

n^{\text{th}}

term of an AP is given by

T_n = a + (n-1)d

Apply the consecutive terms property of a GP

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

For three consecutive terms of a GP, the square of the middle term equals the product of the first and third terms.

Simplify the quadratic equation to find the relationship between

a

and

d

a^2 + 8ad + 16d^2 = a^2 + 14ad + 13d^2 \implies 3d^2 = 6ad \implies d = 2a

Since the AP is non-constant,

d \neq 0

, allowing division by

3d

Formulate the sum of the first

6

terms of the AP

S_6 = 3(2a + 5d) = 72 \implies 2a + 5d = 24

The sum of the first

n

terms of an AP is

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

Substitute

2a = d

into the linear equation and solve for

d

d + 5d = 24 \implies 6d = 24 \implies d = 4

Replacing

2a

with

d

reduces the equation to a single variable.

Anahtar Kavram

Relating non-consecutive terms of an Arithmetic Progression to form a Geometric Progression

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