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Question

How many terms of the AP 78, 71,64..... are needed to give the sum 465? Also find the last term of this AP.

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Solution

Sn=n/2 {2a+(n-1) d}
where Sn is sum of nterm in ap
a is first term and d is common difference
465=n/2{78.2+(n-1)(-7)}
930=n (156-7n+7)
930=163n-7n^2
7n^2-163n+930=0
now use quadratic formula for finding roots
n=10,186/14
hence only n=10 possible
t10=78-9x7=15

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