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  • If the sum of the first 6 terms of an A.P. is 36 and that of the first  16 terms is 256, find the sum of the first 11 terms.   

If the sum of the first 6 terms of an A.P. is 36 and that of the first  16 terms is 256, find the sum of the first 11 terms. 

 

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If the sum of the first 6 terms of an A.P. is 36 and that of the first  16 terms is 256, find the sum of the first 11 terms. 

Sum of n terms in AP S_n=\frac{n}{2}\left [ 2a+(n-1)d \right ]

According to the statements given,

Sum of first six terms S_6=\frac{6}{2}\left [ 2a+(6-1)d \right ]

3\left [ 2a+5d \right ]=36

\left [ 2a+5d \right ]=12........(1)

Sum of first fourteen terms S_{16}=\frac{16}{2}\left [ 2a+(16-1)d \right ]

8\left [ 2a+15d \right ]=256

\left [ 2a+15d \right ]=32........(2)

Subtracting eqn(2) from eqn(1)

2a+5d-2a-15d=12-32

-10d=-20

d=2.

Putting d in eqn (1)

2a+5(2)=12

2a+10=12

2a=2

a=1

Sum of n numbers of AP

S_{11}=\frac{11}{2}\left [ 1\times 2+(11-1)2 \right ]

S_{11}=121

 

Posted by

Safeer PP

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