Question

Prove that sum of $\left( p+q \right)$ terms is $\dfrac{p+q}{2}\left[ a+b+\dfrac{a-b}{p-q} \right]$. The ${{p}^{th}}$ term of an AP is ‘a’ and ${{q}^{th}}$ term is ‘b’.

Answer 1

Hint: Assume that the first term of AP is ‘x’ and the common difference is ‘y’. Use the fact that the ${{n}^{th}}$ term of AP is given by ${{a}_{n}}=x+\left( n-1 \right)y$, where ${{a}_{n}}$ is the ${{n}^{th}}$ term, ‘x’ is the first term and ‘y’ is the common difference. Simplify the equations to eliminate variables ‘x’ and ‘y’. Use the formula ${{S}_{n}}=\dfrac{n}{2}\left[ 2x+\left( n-1 \right)y \right]$ for calculating the sum of first ‘n’ terms.

Complete step-by-step answer:
We know that ${{p}^{th}}$ term of an AP is ‘a’ and ${{q}^{th}}$ term is ‘b’. We have to prove that sum of $\left( p+q \right)$ terms is $\dfrac{p+q}{2}\left[ a+b+\dfrac{a-b}{p-q} \right]$.
Let’s assume that the first term of AP is ‘x’ and the common difference is ‘y’.
We know that that the ${{n}^{th}}$ term of AP is given by ${{a}_{n}}=x+\left( n-1 \right)y$, where ${{a}_{n}}$ is the ${{n}^{th}}$ term, ‘x’ is the first term and ‘y’ is the common difference.
Substituting $n=p$ in the above expression, we have ${{a}_{p}}=x+\left( p-1 \right)y$.
We know that ${{p}^{th}}$ term is ‘a’. Thus, we have $a=x+\left( p-1 \right)y.....\left( 1 \right)$.
Substituting $n=q$ in the above expression, we have ${{a}_{q}}=x+\left( q-1 \right)y$.
We know that ${{q}^{th}}$ term is ‘b’. Thus, we have $b=x+\left( q-1 \right)y.....\left( 2 \right)$.
We will now solve equations (1) and (2). Subtracting equation (2) from equation (1), we have $a-b=x+\left( p-1 \right)y-\left( x+\left( q-1 \right)y \right)$.
Simplifying the above equation, we have $a-b=y\left( p-q \right)$.
Rearranging the terms of the above equation, we have $y=\dfrac{a-b}{p-q}.....\left( 3 \right)$.
Substituting equation (3) in equation (1), we have $a=x+\left( p-1 \right)\left( \dfrac{a-b}{p-q} \right)$.
Simplifying the above equation by rearranging the terms, we have $x=a-\left( p-1 \right)\left( \dfrac{a-b}{p-q} \right)$.
Solving the above equation by taking LCM, we have $x=\dfrac{a\left( p-q \right)-\left( p-1 \right)\left( a-b \right)}{p-q}=\dfrac{ap-aq-ap+bp+a-b}{p-q}$.
Further simplifying the above expression, we have $x=\dfrac{bp-b-aq+a}{p-q}=\dfrac{b\left( p-1 \right)-a\left( q-1 \right)}{p-q}.....\left( 4 \right)$.
So, the first term of the given AP is $x=\dfrac{b\left( p-1 \right)-a\left( q-1 \right)}{p-q}$ and the common difference is $y=\dfrac{a-b}{p-q}$.
We will now calculate the sum of $\left( p+q \right)$ terms of the given AP.
We know that the formula for the sum of ‘n’ terms is ${{S}_{n}}=\dfrac{n}{2}\left[ 2x+\left( n-1 \right)y \right]$.
Substituting $n=p+q,y=\dfrac{a-b}{p-q},x=\dfrac{b\left( p-1 \right)-a\left( q-1 \right)}{p-q}$ in the above formula, we have ${{S}_{p+q}}=\dfrac{p+q}{2}\left[ 2\dfrac{b\left( p-1 \right)-a\left( q-1 \right)}{p-q}+\left( p+q-1 \right)\left( \dfrac{a-b}{p-q} \right) \right]$.
Simplifying the above expression, we have ${{S}_{p+q}}=\dfrac{p+q}{2}\left[ \dfrac{2bp-2b-2aq+2a}{p-q}+\dfrac{ap+aq-a-bp-bq+b}{p-q} \right]$.
Thus, we have ${{S}_{p+q}}=\dfrac{p+q}{2}\left[ \dfrac{bp-b-aq+a+ap-bq}{p-q} \right]$.
Taking out the common terms, we have ${{S}_{p+q}}=\dfrac{p+q}{2}\left[ \dfrac{b\left( p-q \right)+a\left( p-q \right)+a-b}{p-q} \right]$.
Thus, we have ${{S}_{p+q}}=\dfrac{p+q}{2}\left[ a+b+\dfrac{a-b}{p-q} \right]$.
Hence, we have proved that the sum of $\left( p+q \right)$ terms of the given AP is ${{S}_{p+q}}=\dfrac{p+q}{2}\left[ a+b+\dfrac{a-b}{p-q} \right]$.

Note: We can also solve this question by calculating the ${{\left( p+q \right)}^{th}}$ term of the given AP and then calculate the sum using the formula ${{S}_{n}}=\dfrac{n}{2}\left( a+{{a}_{n}} \right)$. We must know that Arithmetic Progression is a sequence of numbers such that the difference between two consecutive terms is a constant.