Question

how many terms of the arithmetic progression 25,22,19 are needed to give the sum of 116 also find the last term. ​

Answer 1

Answer:

We know the formula that sn= n/2 [2a +(n-1) d]

Sn= 116

a= 25

d=22-25 =-3

n=?

116= n/2 [2×25 +(n-1) -3]

232= n[50-3n+3]

232=53n-3n^2

3n^2-53n+232= 0

3n^2-24n-29n+232=0

3n(n-8)-29(n-8)=0

(n-8) (3n-29) =0

n=8

Hence the number of terms of AP is 8.

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