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 = ${\displaystyle \frac{n}{2}[2a+(n-1)d]}$

⇒ S_n = ${\displaystyle \frac{n}{2}[2\times 1+(n-1)\times (-3)]}$

⇒ S_n = ${\displaystyle \frac{n}{2}(2-3n+3)}$

⇒ S_n = ${\displaystyle \frac{n}{2}(5-3n)}$   ...(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

S_n = ${\displaystyle \frac{80}{2}[5-3\times 80]}$

= 40(5 – 240)

= 40 × (–235)

= –9400

Hence, the required sum is –9400.