How many terms of the AP 26,21,16,11,... are needed to give the sum of 11?
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'a' or the first term=26
and 'd' or the common difference =21-26=-5
n=?
Sum of n terms of A.P=11
Sum of n terms of A.P=n/2(2a+(n-1)d
11=n/2(52-5n+5)
22=n(57-5n)
22=57n-5n2
5n2-57n+22=0
By splitting the middle term,
5n2-55n-2n+22=0
5n(n-11)-2(n-11)=0
(5n-2)(n-11)=0
n-11=0
n=11
Therefore 11 terms are needed to get the sum 11
and 'd' or the common difference =21-26=-5
n=?
Sum of n terms of A.P=11
Sum of n terms of A.P=n/2(2a+(n-1)d
11=n/2(52-5n+5)
22=n(57-5n)
22=57n-5n2
5n2-57n+22=0
By splitting the middle term,
5n2-55n-2n+22=0
5n(n-11)-2(n-11)=0
(5n-2)(n-11)=0
n-11=0
n=11
Therefore 11 terms are needed to get the sum 11
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