Question

If the sum of n terms of an A.P is $nA+{{n}^{2}}B$, where A, B are constants, then its common difference will be
(a) A – B
(b) A + B
(c) 2A
(d) 2B

Answer 1

Hint: First of all write the sum of n terms of AP as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Now, separate the terms and write them in terms of n and ${{n}^{2}}$. Now, compare this sum with the given sum $nA+{{n}^{2}}B$ and find the values of A and B in terms of a and d and from here find the value of the common difference.
Complete step-by-step answer:
We are given that the sum of n terms of an A.P is equal to $nA+{{n}^{2}}B$, where A and B are constants. We have to find the value of its common difference. Before proceeding with the question, let us see what an Arithmetic Progression or AP is.
Arithmetic Progression (AP): It is a mathematical sequence in which the difference between two consecutive terms is always a constant.
For example 2, 4, 6, 8, 10…. are in AP with common difference 2.
The ${n}^{th}$ term of AP is given by ${{a}_{n}}=a+\left( n-1 \right)d$ and the sum of n terms of AP is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, where a is the first term and d is the common difference of AP.
Now, we know that the sum of n terms of AP is given by
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
We can further simplify it as follows:
${{S}_{n}}=\dfrac{n}{2}\left( 2a \right)+\dfrac{n}{2}\left( n-1 \right)d$
${{S}_{n}}=na+\dfrac{{{n}^{2}}}{2}d-\dfrac{n}{2}d$
${{S}_{n}}=n\left( a-\dfrac{d}{2} \right)+{{n}^{2}}\left( \dfrac{d}{2} \right)....\left( i \right)$
Also, we are given that the sum of n terms of AP is
${{S}_{n}}=nA+{{n}^{2}}B.....\left( ii \right)$
By equating ${{S}_{n}}$ from equation (i) and (ii), we get,
$n\left( a-\dfrac{d}{2} \right)+{{n}^{2}}\left( \dfrac{d}{2} \right)=nA+{{n}^{2}}B$
We are given that A and B are constants and we already know that a and d are the first term and common difference that also remains constant for an AP. So, by comparing the coefficients of the terms in the LHS and RHS in the above equation, we get,
$\left( a-\dfrac{d}{2} \right)=A....\left( iii \right)$
$\left( \dfrac{d}{2} \right)=B....\left( iv \right)$
From the equation, (iv), we have
$\dfrac{d}{2}=B$
By multiplying 2 on both the sides, we get,
$2\left( \dfrac{d}{2} \right)=2B$
$d=2B$
So, we get the value of the common difference as 2B.
Hence, option (d) is the right answer.

Note: In this question, many students make this mistake of writing the expression of the sum of n term in terms of A and B like $\left( n \right)a+\left[ \dfrac{n}{2}\left( n-1 \right) \right]d$ which is wrong because we need to express it in terms of n and ${{n}^{2}}$so that we can compare it with the given sum. So, this must be there in mind. Also, students must note that A present in the given expression that is $nA+{{n}^{2}}B$ is not the first term of AP but is an arbitrary constant. So, this must be taken care of.