Question

If k + 3, 2k + 1, k + 7 are in A.P., then find this progression up to 5 terms.

Answer 1

Given the three terms are in A.P.:

k + 3, 2k + 1, k + 7

For terms in AP.

(2k + 1) − (k + 3) = (k + 7) − (2k + 1)

k − 2 = 6 − k

2k = 8

k = `8/2`

k = 4

a₁ = k + 3

a₁ = 4 + 3

a₁ = 7

a₂ = 2k + 1

a₂ = 2(4) + 1

a₂ = 8 + 1

a₂ = 9

a₃ = k + 7

a₃ = 4 + 7

a₃ = 11

Common difference d = 9 − 7

= 2

a₄ = a + (n − 1)d

= 7 + (4 − 1)2

= 7 + 3(2)

= 7 + 6

= 13

a₅ = a + (n − 1)d

= 7 + (5 − 1)2

= 7 + 4(2)

= 7 + 8

= 15