How many terms of the ap 25 28 31 34....Are needed to give you a the sum 1070?
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a=25
d=28-25=3
Sn=1070
Using Sn=n/2(2a+(n-1)d)
1070=n/2(2×25+(n-1)3)
1070=n/2(50+(n-1)3)
1070=n/2(50+3n-3)
1070×2=n(50+3n-3)
2140=n(50-3+3n)
2140=n(47+3n)
2140=47n+3n²
0= -2140+47n+3n²
3n²+47n-2140=0
3n²+60n-107n-2140=0
3n(n+20)-107(n+20)=0
(3n-107)(n+20)=0
3n=107
n=107/3
n=35.6
n= -20
So as we know that in an AP the terms can't be negative
Therefore, 35.6 roundoff 36 is the answer
The terms required=36
d=28-25=3
Sn=1070
Using Sn=n/2(2a+(n-1)d)
1070=n/2(2×25+(n-1)3)
1070=n/2(50+(n-1)3)
1070=n/2(50+3n-3)
1070×2=n(50+3n-3)
2140=n(50-3+3n)
2140=n(47+3n)
2140=47n+3n²
0= -2140+47n+3n²
3n²+47n-2140=0
3n²+60n-107n-2140=0
3n(n+20)-107(n+20)=0
(3n-107)(n+20)=0
3n=107
n=107/3
n=35.6
n= -20
So as we know that in an AP the terms can't be negative
Therefore, 35.6 roundoff 36 is the answer
The terms required=36
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