Question

The pth term of an A.P is a and qth term is b. Prove that the sum of its (p+q) term is (p+q) /2 [a+b+(a-b)/p-q)].

Answer 1

Detailed solution in the photo

Answer 2

Answer:

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

Step-by-step explanation:

The nth terms of an AP is

$a_n=f+(n-1)d$

Where f is first term and d is common difference.

It is given that pth term of an A.P is a and qth term is b.

$a_p=f+(p-1)d=a$                .... (1)

$a_q=f+(q-1)d=b$             ..... (2)

Subtract equation (2) from equation (1).

$(p-1)d-(q-1)d=a-b$

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

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

$d=\frac{a-b}{p-q}$

The sum of nth term of an AP is

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

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

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

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

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

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

Hence proved.