Question

How many terms of the following series may be taken so that their sum is 66? -9,-6,3,…

Answer 1

Step-by-step explanation:

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Answer 2

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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