Question

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

Answer 1

The given AP is 5, 12, 19,.......

Here, a = 5, d = 12 – 5 = 7 and n = 50.

Since there are 50 terms in the AP, so the last term of the AP is a₅₀.

l = a₅₀ = 5 + (50 – 1) × 7   ...[a_n = a + (n – 1)d]

= 5 + 343

= 348

Thus, the last term of the AP is 348.

Now,

Sum of the last 15 terms of the AP

= S₅₀ – S₃₅ 

= `50/2 [2 xx 5 + (50 - 1) xx 7] - 35/2 [2 xx 5 + (35 - 1) xx 7]`   ...`{S_n = n/2 [2a + (n - 1)d]}`

= `50/2 xx (10 + 343) - 35/2 xx (10 + 238)`

= `50/2 xx 353 - 35/2 xx 248`

= `(17650 - 8680)/2`

= `8970/2`

= 4485

Hence, the required sum is 4485.