Question

Given that the ${{\left( p+q \right)}^{th}}$ and ${{\left( p-q \right)}^{th}}$ terms of an AP are m and n then, the value of ${{p}^{th}}$ term is
\[\begin{align}
  & A.\dfrac{1}{2}\left( m+n \right) \\
 & B.\sqrt{mn} \\
 & C.m+n \\
 & D.mn \\
\end{align}\]

Answer 1

Hint: To solve this question, we will use the fact that ${{n}^{th}}$ term of AP is given as ${{a}_{n}}=a+\left( n-1 \right)d$ where ‘a’ is first term and d is common difference. First, we will write this formula for ${{\left( p+q \right)}^{th}}$ and ${{\left( p-q \right)}^{th}}$ and then subtract both. Finally, in the obtained equation we will try to obtain the value of d. Then, we will substitute the value of d obtained in one of the equations to get the result.

Complete step by step answer:
Given that, ${{\left( p+q \right)}^{th}}$ term of AP = m and ${{\left( p-q \right)}^{th}}$ term of AP = n
Let us first understand the ${{n}^{th}}$ term of AP.
Arithmetic progression or AP is a sequence of numbers such that the difference of any two successive members is a constant.
If ${{n}^{th}}$ term of an AP is ${{a}_{n}}$ then it can be calculated by using formula:
\[{{a}_{n}}=a+\left( n-1 \right)d\]
Where, a = first term, d = common difference, n = number of terms.
Given, ${{\left( p+q \right)}^{th}}$ term of AP is m
Using this formula stated above we get:
\[{{\left( p+q \right)}^{th}}\text{ term}=m=a+\left( p+q-1 \right)d\]
As here, \[n=p+q\text{ and }{{\left( p+q \right)}^{th}}\text{ term}=m\]
Let ‘a’ be first term of AP with common distance d.
\[m=a+\left( p+q-1 \right)d\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
Similarly, consider ${{\left( p-q \right)}^{th}}$ term as n.
Using the formula stated above for ${{\left( p-q \right)}^{th}}$ term we get:
\[{{\left( p-q \right)}^{th}}\text{ term}=n=a+\left( p-q-1 \right)d\]
Where, AP has the first term ‘a’ and common difference d.
Here, we only have one AP, so the first term is always ‘a’ and the common difference is always ‘d’.
\[n=a+\left( p-q-1 \right)d\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}\]
Subtracting equation (i) and (ii) we get:
\[\begin{align}
  & m-n=a+\left( p+q-1 \right)d-a-\left( p-q-1 \right)d \\
 & m-n=a-a+pd+qd-d-pd+qd+d \\
 & m-n=2qd \\
 & d=\dfrac{m-n}{2q} \\
\end{align}\]
Putting this value of d in equation (i):
\[m=a+\left( p+q-1 \right)\dfrac{\left( m-n \right)}{2q}\]
Separating (p-1) and q term in RHS of above equation:
\[m=a+\dfrac{\left( p-1 \right)\left( m-n \right)}{2q}+q\dfrac{\left( m-n \right)}{2q}\]
Cancelling q we get:
\[m=a+\dfrac{\left( p-1 \right)\left( m-n \right)}{2q}+\dfrac{\left( m-n \right)}{2}\]
Substituting $d=\dfrac{m-n}{2q}$ again in above we get:
\[m=a+\left( p-1 \right)d+\dfrac{\left( m-n \right)}{2}\]
Taking ‘m’ to other side:
\[\begin{align}
  & a+\left( p-1 \right)d=m-\dfrac{\left( m-n \right)}{2} \\
 & a+\left( p-1 \right)d=\dfrac{2m-m+n}{2} \\
 & a+\left( p-1 \right)d=\dfrac{n+m}{2} \\
\end{align}\]
The LHS of above question is the ${{p}^{th}}$ term is $a+\left( p-1 \right)d$
Therefore, the value of ${{p}^{th}}$ term is $\dfrac{m+n}{2}$

So, the correct answer is “Option A”.

Note: Students might get confused at the point where we first used $d=\dfrac{m-n}{2q}$ in a equation and then again substituted d as $\dfrac{m-n}{2q}$ in the same question. This step was done to eliminate 'q' obtained in the equation. Observe that, we had to calculate ${{p}^{th}}$ term of AP which is of the type $a+\left( p-1 \right)d$ where ‘a’ is first term and ‘d’ is common difference. So, this question doesn't contain 'q' in it, therefore we had to remove it. Also, we again substituted $\dfrac{m-n}{2q}$ as d because $a+\left( p-1 \right)d$ had 'd' in it.