Question

Find the sum of last ten terms of the a.p.:8,10,12,14....,126.

Answer 1

series from end=126,...........14,12,10,8
so,a=126
d=12-14=-2
sum of first 10 no. =(10/2){2×126+(10-1)×(-2)}
=5(252+9×(-2))
=5(252-18)
=5(234)
=1170
so sum of last 10 terms of a.p. 8,10,12,14,........126=1170

Answer 2

HEY MATE HERE IS YOUR ANSWER

Given A.P=8,10,12....A.P=8,10,12....

n=60n=60

a=8a=8

d=2d=2

an=a+(n−1)dan=a+(n−1)d

a60=8+59×2a60=8+59×2

=126=126

Hence sum of the last 10 items is

d=−2,a=126,r=10d=−2,a=126,r=10

s=n2s=n2[2a+(n−1)d][2a+(n−1)d]

S=102S=102[2×126+(10−1)(−2)][2×126+(10−1)(−2)]

=5(252−18)=5(252−18)

=1170=1170

Hence the sum of last to terms =1170

HOPE THIS HELPS YOU...