Question

The $p^{th}$ and $q^{th}$ terms an $A.P$ are respectively $a$ and $b.$ Then sum of $(p+q)$ terms is

• A. $\frac{p+q}{2}(a+b+\frac{a-b}{p-q})$
• B. $\frac{p+q}{2}(a-b-\frac{a+b}{p+q})$
• C. $(p+q)(a+b+\frac{p-q}{a-b})$
• D. $\frac{p+q}{2}(a+b+\frac{p-q}{a-b})$

Answer 1

Answer: (A)

$\frac{p+q}{2}(a+b+\frac{a-b}{p-q})$

Let the first term of the A.P be '$c$' and its common difference be '$d$'.

So its $p^{th}$ term is $c+(p-1)d=a$ $...\left(1\right)$

and its $q^{th}$ term is $c+(q-1)d=b$ $...\left(2\right)$

Now, add equation $\left(1\right)$ and $\left(2\right)$; we get

$\Rightarrow 2c+(p+q-2)d=a+b$

$\Rightarrow 2c+(p+q-1)d-d=a+b$

$\Rightarrow 2c+(p+q-1)d=a+b+d$ $...\left(3\right)$

Now, subtracting $\left(1\right)$ from $\left(2\right)$; we get

$\Rightarrow (p-q)d=a-b$

$\Rightarrow d=\frac{(a-b)}{(p-q)}$ $...\left(4\right)$

Now, the sum of $(p+q)$ terms is

$S_{(p+q)}=\frac{(p+q)}{2}[2c+(p+q-1\left)d\right]$ $...\left(5\right)$

Substitute equation $\left(3\right)$ and $\left(4\right)$ in $\left(5\right)$; we get

$S_{(p+q)}=\frac{(p+q)}{2}[a+b+d]$

$S_{(p+q)}=\frac{(p+q)}{2}[a+b+\frac{(a-b)}{(p-q)}]$