Question

If $S_n=nP + \dfrac{1}{2}n(n - 1)Q$, where $S_n$​ denotes the sum of the first n terms of an A.P., then the common difference is
A) P+Q
B) 2Q
C) Q
D) 2P+3Q

Answer 1

A progression is a sequence of numbers that follow a specific pattern. Here, we are given a sum of n terms where all the terms are in AP. Then, we will substitute the values for n = 1 and then n = 2. After that, we will find the first term and the second term. Thus, we will find the common difference (d) which is equal to the second term – first term. Also, we can find using another method too by comparing the given sum of n terms with the formula of AP (i.e. ${S_n} = \dfrac{n}{2}(2a + (n - 1)d)$ ), where a is the first term, d is the common difference and ${S_n}$ is the sum of the first n terms. We will get the first term and the common difference too.

Complete step by step solution:
Let ${a_1},{a_2},....,{a_n}$ be the given AP.

Also given that,
Sum of n terms is $nP + \dfrac{1}{2}n(n - 1)Q$
$\therefore {S_n} = nP + \dfrac{1}{2}n(n - 1)Q$ --------- (1)

First,
Substituting the value n = 1 in equation (1), we will get,
$\Rightarrow {S_1} = (1)P + \dfrac{1}{2}1(1 - 1)Q$
On evaluating this above equation, we will get,
$\Rightarrow {S_1} = P + \dfrac{1}{2}(0)Q$
$\Rightarrow {S_1} = P + 0$
$\Rightarrow {S_1} = P$

But, sum of the first terms will be the first term
i.e. ${a_1} = {S_1} = P$

Next,
Substituting the value n = 2 in equation (1), we will get,
$\Rightarrow {S_2} = (2)P + \dfrac{1}{2}2(2 - 1)Q$
On evaluating this above equation, we will get,
$\Rightarrow {S_2} = 2P + \dfrac{1}{2}2(1)Q$
$\Rightarrow {S_2} = 2P + Q$

Thus, the sum of the first two terms is first term + second term
$\therefore {S_2} = {a_1} + {a_2}$
$\Rightarrow {S_2} - {a_1} = {a_2}$
$\Rightarrow {a_2} = {S_2} - {a_1}$
Substituting the values of ${a_1} = P$ and ${S_2} = 2P + Q$, we will get,
$\Rightarrow {a_2} = 2P + Q - P$
$\Rightarrow {a_2} = P + Q$

Thus, the common difference is
$d = {a_2} - {a_1}$
Substituting the values, we will get,
$\Rightarrow d = (P + Q) - P$
$\Rightarrow d = P + Q - P$
$\Rightarrow d = Q$
Therefore, the common difference is $Q$.

Additional information:
There are three different types of progressions. They are:

1. Arithmetic Progression (AP)
2. Geometric Progression (GP)
3. Harmonic Progression (HP)
   Arithmetic Progression (AP) is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value. An arithmetic progression is a sequence where each term, except the first term, is obtained by adding a fixed number to its previous term.

Note:
Alternative approach:
Let ‘a’ be the first term and d be the common difference of an AP.
$\therefore {S_n} = \dfrac{n}{2}(2a + (n - 1)d)$
$\Rightarrow {S_n} = \dfrac{{2an}}{2} + \dfrac{{n(n - 1)d}}{2}$
$\Rightarrow {S_n} = an + \dfrac{1}{2}n(n - 1)d$ -------- (i)

Also given that, sum of n terms is
$\therefore {S_n} = nP + \dfrac{1}{2}n(n - 1)Q$ --------- (ii)
Comparing the coefficients of the respective powers of n, of equations (i) and (ii), we will get,
a = P and d = Q
Thus, the first term of AP is P and the common difference is Q.
Hence, the common difference of the given AP is Q.