Question

Show that a₁, a_2,..., a_n... form an AP where a_n is defined as below:

a_n = 3 + 4n

Also, find the sum of the first 15 terms.

Answer 1

a_n = 3 + 4n

a₁ = 3 + 4(1) = 7

a₂ = 3 + 4(2) = 3 + 8 = 11

a₃ = 3 + 4(3) = 3 + 12 = 15

a₄ = 3 + 4(4) = 3 + 16 = 19

It can be observed that

a₂ − a₁ = 11 − 7 = 4

a₃ − a₂ = 15 − 11 = 4

a₄ − a₃ = 19 − 15 = 4

i.e., a_{k + 1} − a_k is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.

`S_n = n/2 [2a + (n - 1)d]`

`S_15 = 15/2 [2(7) + (15 - 1) × 4]`

= `15/2 [(14) + 56]`

= `15/2 (70)`

= 15 × 35

= 525