If the sum of n terms of an A.P. is \(nP+\frac{1}{2}n(n-1)Q\), where P and Q are constants, find the common difference.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Q
The sum of the first \(n\) terms already contains every term, so the first term is \(S_1\) and each later term is \(a_n=S_n-S_{n-1}\). The common difference is the second term minus the first.
The given sum, with the \(\frac{1}{2}\) multiplying \(n(n-1)Q\): \(S_n=nP+\frac{1}{2}n(n-1)Q\)
First term \(a_1=S_1=P+0=P\)
Sum of the first two terms: \(S_2=2P+\frac{1}{2}\times2\times1\times Q=2P+Q\)
Second term \(a_2=S_2-S_1=(2P+Q)-P=P+Q\)
Common difference \(d=a_2-a_1=(P+Q)-P=Q\)
Shortcut: The sum of an A.P. has the standard form \(S_n=na+\frac{1}{2}n(n-1)d\), so the given sum can be compared with it term by term.
\(nP\) stands in the place of \(na\), so \(a=P\)
\(\frac{1}{2}n(n-1)Q\) stands in the place of \(\frac{1}{2}n(n-1)d\), so \(d=Q\)