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The sum of the first n terms of an A. P. is 2n and the sum of the first 2n terms is n. Find the sum of the first 4n terms.

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User Oasia
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1 Answer

3 votes

Answer:

The sum of first 4n terms is -10n.

Explanation:

The formula for sum of n terms of an AP is


S_n=(n)/(2)[2a+(n-1)d]

It is given that the sum of the first n terms of an A. P. is 2n and the sum of the first 2n terms is n.


(n)/(2)[2a+(n-1)d]=2n


2a+(n-1)d=4 ..... (1)


(2n)/(2)[2a+(2n-1)d]=n


2a+(2n-1)d=1 ..... (2)

Solve equation (1) and (2) by elimination method.


d=-(3)/(n)


a=(1)/(2)(7-(3)/(n))

The sum of first 4n terms is


S_(4n)=(4n)/(2)[2a+(4n-1)d]


S_(4n)=2n[2a+(4n-1)d]

Put the value of a and d.


S_(4n)=2n[2((1)/(2)(7-(3)/(n)))+(4n-1)(-(3)/(n))]


S_(4n)=2n[7-(3)/(n)-12+(3)/(n)]


S_(4n)=2n[-5]


S_(4n)=-10n

Therefore the sum of 4n terms is -10n.

answered
User Stevecowling
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8.0k points

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