How many terms of the ap 1 4 7 are needed to give the sum 715?
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Sn=n/2(2a+(n-1)d)
715=n/2(2×1+(n-1)3)
715×2=n(2+3n-3)
1430=n(3n-1)
1430=3n^2-n
3n^2-n-1430=0
3n^2-66n+65n-1430=0
3n(n-22)+65(n-22)=0
(n-22)(3n+65)=0
n-22=0. 3n+65=0
n=22. n=-65/3
n can not be in fraction hence, n=-65/3 is not possible
hence,n=22 is the answer
hope it helps...
715=n/2(2×1+(n-1)3)
715×2=n(2+3n-3)
1430=n(3n-1)
1430=3n^2-n
3n^2-n-1430=0
3n^2-66n+65n-1430=0
3n(n-22)+65(n-22)=0
(n-22)(3n+65)=0
n-22=0. 3n+65=0
n=22. n=-65/3
n can not be in fraction hence, n=-65/3 is not possible
hence,n=22 is the answer
hope it helps...
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