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

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

#### Solution

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.

Concept: Arithmetic Progression
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