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If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of it's first n terms.

 

 
 
 
 
 

Answers (1)

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 four terms S_4=\frac{4}{2}\left [ 2a+(4-1)d \right ]

2\left [ 2a+3d \right ]=40

\left [ 2a+3d \right ]=20........(1)

Sum of first fourteen terms S_{14}=\frac{14}{2}\left [ 2a+(14-1)d \right ]

7\left [ 2a+13d \right ]=280

\left [ 2a+13d \right ]=40........(2)

Subtracting eqn(2) from eqn(1)

2a+3d-2a-13d=20-40

-10d=-20

d=2.

Putting d in eqn (1)

2a+3(2)=20

2a+6=20

2a=14

a=7

Sum of n numbers of AP

S_n=\frac{n}{2}\left [ 7\times 2+(n-1)2 \right ]

S_n=n\left [ 7+n-1 \right ]

S_n=n\left [ 6+n \right ]

S_n=n^2+6n

 

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Safeer PP

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