The first 3 terms of an Arithmetic Progression are x+3, 2x+6 and 8.

Question

Find the value of x and the sum of the first 12 terms. How is it done?

Answer 1

The first three terms in arithmetic progression are x +3 , 2x + 6, 8

Common difference is equal.

(2x + 6) - ( x + 3) = 8 - (2x + 6)

2x + 6 - x - 3 = 8 - 2x - 6

x + 3 = - 2x + 2

3x = -1

x = -1/3

First term is (-1/3) + 3 = ( -1 + 9)/3 = 8/3

a = 8/3

Second term = 2(-1/3) + 6 = (-2/3) + 6 = ( - 2 + 18)/3 = 16/3

d = t₂ - t₁ = 16/3 - 8/3 = 8/3

d = 8/3

Apply the formula Sn = n/2[2a +(n - 1)d]

n = 12, a = 8/3, d = 8/3

S12 = 6[(16/3) +(11)8/3)

= 6(16 + 88)/3) 

= 2(104)

= 208

Sum of 12 terms in the progression is 208.