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Question: Given that the \({{\left( p+q \right)}^{th}}\) and \({{\left( p-q \right)}^{th}}\) terms of an AP ar...

Given that the (p+q)th{{\left( p+q \right)}^{th}} and (p−q)th{{\left( p-q \right)}^{th}} terms of an AP are m and n then, the value of pth{{p}^{th}} term is

& A.\dfrac{1}{2}\left( m+n \right) \\\ & B.\sqrt{mn} \\\ & C.m+n \\\ & D.mn \\\ \end{aligned}$$
Explanation

Solution

To solve this question, we will use the fact that nth{{n}^{th}} term of AP is given as an=a+(n−1)d{{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 (p+q)th{{\left( p+q \right)}^{th}} and (p−q)th{{\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, (p+q)th{{\left( p+q \right)}^{th}} term of AP = m and (p−q)th{{\left( p-q \right)}^{th}} term of AP = n
Let us first understand the nth{{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 nth{{n}^{th}} term of an AP is an{{a}_{n}} then it can be calculated by using formula:
an=a+(n−1)d{{a}_{n}}=a+\left( n-1 \right)d
Where, a = first term, d = common difference, n = number of terms.
Given, (p+q)th{{\left( p+q \right)}^{th}} term of AP is m
Using this formula stated above we get:
(p+q)th term=m=a+(p+q−1)d{{\left( p+q \right)}^{th}}\text{ term}=m=a+\left( p+q-1 \right)d
As here, n=p+q and (p+q)th term=mn=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+(p+q−1)d . . . . . . . . . . (i)m=a+\left( p+q-1 \right)d\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}
Similarly, consider (p−q)th{{\left( p-q \right)}^{th}} term as n.
Using the formula stated above for (p−q)th{{\left( p-q \right)}^{th}} term we get:
(p−q)th term=n=a+(p−q−1)d{{\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+(p−q−1)d . . . . . . . . . . (ii)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:

& 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{aligned}$$ Putting this value of d in equation (i): $$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{aligned} & 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{aligned}$$ 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.