Question

there are 10 terms in ap if sum of first 3 terms is 36 and last three term is 141.find sum of last 3 terms.

Answer 1

Answer:

Solution :

a11+a12+a13=141=>3a+33d=141=>a+11d=47

a21+a22+a23=261=>3a+63d=261=>a+21d=87

on solving : a=3

Answer 2

let the a-d,a,a+d,a+2d,a+3d,a+4d,a+5d,a+6d,a+7d,a+8d are the terms of ap

by 1st condition

sum of 1st three terms of an ap is 36

a-d+a+a+d=36

3a=36

a=12.......1

By second condition

sum of last three terms is 141

a+6d+a+7d+a+8d=141

3a+21d=141

divide by 3

a+21d =47

12 +21d=47(from 1)

7d=47-12

7d=35

d=5

the sum of last 3 terms

a+6d=42

a+7d=47

a+8d=52

hope it help u