in an ap if the first term is 22 the common difference is 4 and the sum to n terms is 64 find and answers
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What we should find in this ques
Find n
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we have a=22 and d=4
Sn= n[(2a+( n-1)d]/2
⇒ 64=n(44+4n-4)/2
⇒128= 40n-4n²
⇒ n²-10n + 32=0
solving the quadratic equation we have
n=+_5+_√-28
This is not correct value for n. I think the sum to N term can not be 64 as the terms are 22,26,30 hence we can not get Sn=64 .
Sn= n[(2a+( n-1)d]/2
⇒ 64=n(44+4n-4)/2
⇒128= 40n-4n²
⇒ n²-10n + 32=0
solving the quadratic equation we have
n=+_5+_√-28
This is not correct value for n. I think the sum to N term can not be 64 as the terms are 22,26,30 hence we can not get Sn=64 .
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