Question

Find the sum: 1 + (–2) + (–5) + (–8) + ... + (–236)

Answer 1

Here, first term (a) = 1

And common difference (d) = (–2) – 1 = –3

∵ Sum of n terms of an AP

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

⇒ `S_n = n/2 [2 xx 1 + (n - 1) xx (-3)]`

⇒ `S_n = n/2 (2 - 3n + 3)`

⇒ `S_n = n/2 (5 - 3n)`   .....(i)

We know that

If the last term (l) of an AP is known, then

l = a + (n – 1)d

⇒ – 236 = 1 + (n – 1)(– 3) ......[∵ l = – 236, given]

⇒ – 237 = – (n – 1) × 3

⇒ n – 1 = 79

⇒ n = 80

Now, put the value of n in equation (i), we get on

`S_n = 80/2 [5 - 3 xx 80]`

= `40[5 - 240]`

= `40 xx (-235)`

= – 9400

Here, the required sum is – 9400