# An AP 5, 12, 19,

Here, First term = a = 5

Common difference = d = 12 - 5 = 7

No. of terms = 50

last term will be 50th term.

Using the formula for finding nth term of an A.P.,

l = a50 = a + (50 - 1) × d

l = 5 + (50 - 1) × 7

l = 5 + 343 = 348

Now, sum of last 15 terms = sum of first 50 terms - sum of first 35 terms

i.e. sum of last 15 terms = S50 - S35

Now, Sum of first n terms of an AP is

Sn = [2a + (n - 1)d]

Sum of first 50 terms is given by:

S50 = [2(5) + (50 - 1)(7)]

= 25 × [10 + 343]

= 25 × 353

= 8825

Now, Sum of first 35 terms is given by:

S35 = [2(5) + (35 - 1)(7)]

= (35/2) × [10 + 238]

= (35/2) × 248

= 35 × 124

= 4340

Now, S50 - S35 = 8825 - 4340

= 4485

last term = 348, sum of last 15 terms = 4485

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 