Advertisement Remove all ads

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

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads

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

Advertisement Remove all ads

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
  Is there an error in this question or solution?
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×