Question

The $p^{th}$ term of an A.P is a and $q^{th}$ term is b. Prove that the sum of its (p+q) terms is $\frac{p+q}{2}[a+b+\frac{a-b}{p-q}]$.

Answer 1

As we know that $n^{th}$ term in an A.P. is given as-

$A_n=A+(n-1)D$

Where as $A$ and $D$ be the first term and common difference of the A.P.

Now,

$A_p=a\left(\text{Given}\right)$

$\Rightarrow A+(p-1)D=a.....\left(1\right)$

$A_q=b\left(\text{Given}\right)$

$\Rightarrow A+(q-1)D=b.....\left(2\right)$

Subtracting ${eq}^n\left(2\right)$ from $\left(1\right)$, we have

$(A+(p-1)D)-(A+(q-1)D)=a-b$

$A+(p-1)D-A-(q-1)D=a-b$

$D(p-1-q+1)=a-b$

$\Rightarrow D=\frac{a-b}{p-q}.....\left(3\right)$

Now adding ${eq}^n\left(1\right)\&\left(2\right)$, we have

$(A+(p-1)D)+(A+(q-1)D)=a+b$

$A+(p-1)D+A+(q-1)D=a+b$

$2A+(p-1+q-1)D=a+b$

$2A+(p+q-2)D=a+b$

$2A+(p+q-1)D-D=a+b$

$2A+(p+q-1)D=a+b+D$

$2A+(p+q-1)D=a+b+\frac{a-b}{p-q}.....\left(4\right)(\because D=\frac{a-b}{p-q})$

Now as we know that sum of $n$ terms in an A.P. is given as-

$S_n=\frac{n}{2}[2A+(n-1)D]$

Therefore,

$S_{p+q}=\frac{p+q}{2}[2A+(p+q-1)D]$

$\Rightarrow S_{p+q}=\frac{p+q}{2}(a+b+\frac{a-b}{p-q})(\because \text{From}\left(4\right))$

Hence proved.