The 16th term of an A.p.
is five times its third term. If its 10th term is 41, find the sum of its first fifteen terms.
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Answer:
a+15d=5(a+2d)
a+15d=5a+10d
5d=4a
5/4d=a---‐eq. 1
a+9d=41
5/4d+9d=41
5d+36d/4=41
41d/4=41
d=4
put the value of d in eq 1
a=5
sum of first fifteen terms
sum=n/2[2a+(n-1)d]
=15/2[2×5+(15-1)4]
15/2[10+(14)4]
15/2[66]
15×33
=495
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