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Question 32
How many terms of the AP –15, –13, –11,. … are needed to make the sum –55?

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Solution

Let n be the number of terms needed to make the sum 55.
Here, first term (a) = - 15, common difference (d) = - 13 + 15 = 2
Sum of n terms of an AP, Sn=n2[2a+(n1)d]
55=n2[2(15)+(n1)2] [Sn=55(given)]
55=15n+n(n1)
n216n+55=0
n211n5n+55=0 [by factorization method]
n(n11)5(n11)=0
(n11)(n5)=0
n=5,11
Hence, either 5 or 11 terms are needed to make the sum - 55.

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