Question

An arithmetic progression 5, 12, 19, …. has 50 terms. Find its last term. Hence find the sum of its last 15 terms.

Answer 1

5, 12, 19, …………50 terms
Common difference, d = 7
First term, a = 5

Last term, t₅₀ = a + (50 – 1)d = 5 + (50 – 1) × 7 = 5 + 49 × 7 = 5 + 343 = 348

Sum of last 15 terms = S₅₀ – S₃₅

`=50/2[2xx5+[50-1]xx7]-35/2[2xx5+[35-1]xx7]`

`=25[10+343]-35/2[10+34xx7]`

`=25xx353-35/2xx248`

=8825-4340

=4485

The sum of last 15 terms = 4485.