Find the sum of the series –5+(–8)+(–11)+........+ (–230) ?

–5 + (–8) + (–11) + … + (–230)

Let us find the difference of consecutive terms,

a₂ – a₁ = –8 – (–5) = –3

a₃ – a₂ = –11 – (–8) = –3

As consecutive differences of terms are same, therefore above series is an AP with

First term, a = –5

Common difference, d = –3

Last term, a_n = –230

Let the number of terms be ‘n’.

We know, nth term of an AP is

a_n = a + (n – 1)d

Putting values, we get

–230 = –5 + (n – 1)(–3)

⇒ –225 = (n – 1)(–3)

⇒ n – 1 = 75

⇒ n = 76

We know, the sum of ‘n’ terms of an AP is

Where a_n is nth term.

Putting values, we get

⇒ S₇₆ = 38 × (–235)

⇒ S₇₆ = –8930