How many terms of the following series may be taken so that their sum is 66? -9,-6,3,…
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Let sum of n terms will be 66
Sn=66
Sn=n/2[2a+(n-1)d]
a= -9 and d = -6-(-9)=3
We need to find n
= 66=n/2[2(-9)+(n-1)(3)]
= 66*2=n[-18+3n-3]
= 132=n[-21+3n]
= 132 = -21n+3n^2
= 3n^2-21n-132=0 {Take 3 common}
= n^2-7n-44 =0
= n^2-11n+4n-44=0
= n(n-11)+4(n-11) =0
= (n-11)(n+4)=0
=n= 11,-4
n can't be negative
so 11 terms of this series may be taken so their sum is 66.
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