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Question

Find the minimum number of terms of the AP 64, 60, 56,... so that their sum is 540.

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Solution

We can see that
d=6064=4a=64

We know that ,
Sum of first terms,
Sn=n2(2a+(n1)d)540=n2(2×64+(n1)×(4))
540=64n2n2+2n
n233n+270=0
n215n18n+270=0
n(n15)18(n15)=0
(n15)(n18)=0
n=15,18
Since we need a minimum number of terms therefore n=15

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