how many terms of the ap .24+21+18+15+...be taken continuously so that their sum is -351?
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In Given AP. a = 24
and, d = 21 - 24 = -3
We have to find n for Sn = -351
so, n/2[2(a) + (n-1)d] = -351
so, n[2*24 + (n-1)-3] = -702
= n(48 + 3-3n) = -702
= n(51-3n) = -702
51n - 3n^2 = -702
On Dividing whole Eqn by 3
17n - n^2 = -234
So, n^2 - 17n - 234 = 0
n^2 -26n + 9n -234 = 0
n(n-26) +9(n-26) = 0
so , n= 26,-9
But n can't be negative
So, 26 terms should be taken to get the required sum.
and, d = 21 - 24 = -3
We have to find n for Sn = -351
so, n/2[2(a) + (n-1)d] = -351
so, n[2*24 + (n-1)-3] = -702
= n(48 + 3-3n) = -702
= n(51-3n) = -702
51n - 3n^2 = -702
On Dividing whole Eqn by 3
17n - n^2 = -234
So, n^2 - 17n - 234 = 0
n^2 -26n + 9n -234 = 0
n(n-26) +9(n-26) = 0
so , n= 26,-9
But n can't be negative
So, 26 terms should be taken to get the required sum.
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