Question

The $p^{th}$ term of an A.P. is a and $q^{th}$ term is b.then the sum of its (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. $\frac{p+q}{2}[a+b+\frac{a+b}{p+q}]$
• D. none of these

Answer 1

Answer: (A)

$\frac{p+q}{2}[a+b+\frac{a-b}{p-q}]$

Let the first term of the AP be ${}^{\prime }{c}^{\prime }$ and its common difference be ${}^{\prime }{d}^{\prime }$.

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

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

Adding $\left(1\right)$ & $\left(2\right)$: $2c+(p+q-2)d=a+b$

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

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

Operating $\left(1\right)-\left(2\right)$: $(p-q)d=(a-b)$

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

Sum to $(p+q)$ terms is:

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

Substituting from $\left(3\right)$ and $\left(4\right)$, we get

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

Hence proved