Question

The $p^{\text{th }}$ term of an A.P. is $a$ and $q^{th}$ term is $b$. Then, the sum of its $(p+q)$ terms is

• A. $\frac{p+q}{2}\left[a+b+\frac{a-b}{p+q}\right]$
• B. $\frac{p+q}{2}\left[a+b+\frac{a+b}{p+q}\right]$
• C. $\frac{p+q}{2}\left[a+b+\frac{a-b}{p-q}\right]$
• D. $\frac{p+q}{2}\left[a+b+\frac{a+b}{p-q}\right]$

Answer 1

Answer: (C)

Let $A$ be the first term and $D$ be the common difference of the A.P. . It is given that
$t_p=a\Rightarrow A+(p-1)D=a.....$(i)
$t_q=b\Rightarrow A+(q-1)D=b....$(ii)
Subtracting Eq. (ii) from Eq. (i), we get
$(p-1-q+1)D=a-b$
$\Rightarrow D=\frac{a-b}{p-q}....$(iii)
Adding Eqs. (i) and (ii), we get
$2A+(p+q-2)D=a+b$
$\Rightarrow 2A+(p+q-1)D=a+b+D$
$\Rightarrow 2A+(p+q-1)D=a+b+\frac{a-b}{p-q}.....$(iv)
Now, $S_{p+q}=\frac{p+q}{2}[2A+(p+q-1)D]$
$=\frac{p+q}{2}\left[a+b+\frac{a-b}{p-q}\right]$ [using Eq. (iv)]