Question

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

a_n = 9 − 5n

Also, find the sum of the first 15 terms.

Answer 1

a_n = 9 − 5n

a₁ = 9 − 5 × 1

= 9 − 5

= 4

a₂ = 9 − 5 × 2

= 9 − 10

= −1

a₃ = 9 − 5 × 3

= 9 − 15

= −6

a₄ = 9 − 5 × 4

= 9 − 20

= −11

It can be observed that

a₂ − a₁ = −1 − 4 = −5

a₃ − a₂ = −6 − (−1) = −5

a₄ − a₃ = −11 − (−6) = −5

i.e., a_{k + 1} − a_k is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.

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

`S_15 = 15/2 [2(4) + (15 - 1) (-5)]`

= `15/2 [8 + 14(-5)]`

= `15/2 (8 - 70)`

= `15/2 (-62)`

= 15 × (-31)

= -465