solve for X 1 + 4 + 7 + 10 + .... + X = 287....ap question
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GIVEN ,a=1 ; d=3 ; sum=287
Sum= n/2 [2a+(n-1)d]
=> 287= n/2 [2×1+(n-1)×3]
=> 287=n/2 [2+3n-3]
=> 287×2=n [3n-1]
=> 574=3n×n-n
=> 3n×n-n-574=0
on solving the quadratic equation we get,
n=14
we know that x is the last term..
so,
l= a+(n-1)d
= 1+13×3=1+39=40------->(ANS)
Sum= n/2 [2a+(n-1)d]
=> 287= n/2 [2×1+(n-1)×3]
=> 287=n/2 [2+3n-3]
=> 287×2=n [3n-1]
=> 574=3n×n-n
=> 3n×n-n-574=0
on solving the quadratic equation we get,
n=14
we know that x is the last term..
so,
l= a+(n-1)d
= 1+13×3=1+39=40------->(ANS)
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