If the sum of n terms of an A.P. is nP+21n(n−1)Q, where P and Q are constants, find the common difference.

We have,Sn=nP+21n(n−1)Q⇒Sn−1=(n−1)P+21(n−1)(n−2)QLet an be the nth term. Then,an=Sn−Sn−1⇒an={nP+21n(n−1)Q}−{(n−1)P+21(n−1)(n−2Q}⇒an=P+21(n−1){n−(n−2)}Q⇒an=P+(n−1)Q⇒an−1=P+(n−2)QLet d be the common difference. Then,d=an−an−1={P+(n−1)Q}−{P+(n−2)Q}=Q.

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If the sum of n terms of an A.P. is nP+{1}{2}n(n-1)Q, where P

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If the sum of n terms of an A.P. is $n P + \frac{1}{2} n (n - 1) Q$ , where P and Q are constants, find the common difference.

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If the sum of n terms of an A.P. is

n P + \frac{1}{2} n (n - 1) Q

, where P and Q are constants, find the common difference.

13 mins ago

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13 mins ago

Text solutionVerified

We have,

S_{n} = n P + \frac{1}{2} n (n - 1) Q \Rightarrow S_{n - 1} = (n - 1) P + \frac{1}{2} (n - 1) (n - 2) Q

Let

a_{n}

be the

n^{t h}

term. Then,

a_{n} = S_{n} - S_{n - 1}

\Rightarrow a_{n} = {n P + \frac{1}{2} n (n - 1) Q} - {(n - 1) P + \frac{1}{2} (n - 1) (n - 2 Q}

\Rightarrow a_{n} = P + \frac{1}{2} (n - 1) {n - (n - 2)} Q

\Rightarrow a_{n} = P + (n - 1) Q

\Rightarrow a_{n - 1} = P + (n - 2) Q

Let d be the common difference. Then,

d = a_{n} - a_{n - 1} = {P + (n - 1) Q} - {P + (n - 2) Q} = Q

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Question Text	If the sum of n terms of an A.P. is $n P + \frac{1}{2} n (n - 1) Q$ , where P and Q are constants, find the common difference.
Answer Type	Text solution:1
Upvotes	150

If the sum of n terms of an A.P. is nP+21​n(n−1)Q, where P and Q are constants, find the common difference.

Text solutionVerified

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If the sum of n terms of an A.P. is $n P + \frac{1}{2} n (n - 1) Q$ , where P and Q are constants, find the common difference.