how many terms of ap,2,4,6.... must be taken so that their sum is 156
2003uditraj:
y any one couldn't answer my question
is it so difficult or u ppl have been failed to answer
Chill babe. I will do it.
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Hey sup!
As per the question,
a=2.
d=4-2=2.
S(n)=156.
=>n/2{2a+(n-1)d}=156.
=>n/2{2*2+(n-1)2=156.
=>n/2(4+2n-2)=156.
=>n/2(2+2n)=156.
=>n(2+2n)=156*2.
=>2n+2n^2=312.
=>2n+2n^2-312=0.
=>2(n^2+n-156)=0.
=>n^2+n-156=0/2=0.
=>n^2+13n-12n-156=0.
=>n(n+13)-12(n+13)=0.
=>(n-12)(n+13)=0.
We have got two roots.
We'll discard n+13 as it gives-ve value.
So, n-12=0.
n=12.
12 terms of AP 2,4,6.... must be taken so that their sum is 156.
Hope it helps.
As per the question,
a=2.
d=4-2=2.
S(n)=156.
=>n/2{2a+(n-1)d}=156.
=>n/2{2*2+(n-1)2=156.
=>n/2(4+2n-2)=156.
=>n/2(2+2n)=156.
=>n(2+2n)=156*2.
=>2n+2n^2=312.
=>2n+2n^2-312=0.
=>2(n^2+n-156)=0.
=>n^2+n-156=0/2=0.
=>n^2+13n-12n-156=0.
=>n(n+13)-12(n+13)=0.
=>(n-12)(n+13)=0.
We have got two roots.
We'll discard n+13 as it gives-ve value.
So, n-12=0.
n=12.
12 terms of AP 2,4,6.... must be taken so that their sum is 156.
Hope it helps.
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