how many terms of the arithmetic progression 25,22,19 are needed to gi

1.	how many terms of the arithmetic progression 25,22,19 are needed to give the sum of 116 also find the last term.
Answer» 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. Hope it helps. Thanks Mark it as brainliest.

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

Answer»

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.

Hope it helps.

Thanks

Mark it as brainliest.

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

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