Question

The $p^{\text {th }}$ term of an A.P. is $a$ and $q^{t h}$ 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
$2 A+(p+q-2) D=a+b$
$\Rightarrow 2 A+(p+q-1) D=a+b+D$
$\Rightarrow 2 A +(p+q-1) D=a+b+\frac{a-b}{p-q} .....$(iv)
Now, $ S_{p+q} =\frac{p+q}{2}[2 A+(p+q-1) D] $
$ =\frac{p+q}{2}\left[a+b+\frac{a-b}{p-q}\right]$ [using Eq. (iv)]