Find the sum of last ten terms of the a.p.:8,10,12,14....,126.
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Answered by
34
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
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
madaanpriyanshi:
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Answered by
4
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...
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