Question

If the pth term of an A.P. is x and qth term is y, show that the sum of (p+q) terms ​is

Answer 1

Answer:

Step-by-step explanation:

Given:

p'th term of AP is x; q'th term is y.

Let 'a' be the first term and 'd' the common difference.

p'th term ⇒ a+(p-1)d = x

q'the term⇒ a+(q-1)d = y

x-y = d(p-1) - d(q-1)

x-y = d(p-q)

⇒d= $\frac{x-y}{p-q}$ ⇒ Equation 1.

Sum of 'n' terms of AP = $\frac{n}{2} ( 2a + (n-1)d)$

Sum of (p+q) terms = $\frac{p+q}{2}( 2a+(p+q - 1) d)$

= $\frac{p+q}{2}( a+a+pd+qd-d)$

= $\frac{p+q}{2}( a+a+(p-1)d+qd)$

Adding and subtracting a 'd', and subtituting a+(p-1)d = x;

Sum of (p+q) terms

$=\frac{p+q}{2}( x+a +qd -d + d)\\= \frac{p+q}{2}( x+ a+(q-1)d +d)\\= \frac{p+q}{2}(x+y+d)$

Subtituting for 'd' from Equation 1;

We get

Sum of (p+q) terms = $\frac{p+q}{2}(x+y+\frac{x-y}{p-q} )$

S

Answer 2

Step-by-step explanation:

For future use, note that d=tp−tqp−q=x−yp−q .

Now, a+(p−1)d=x and a+(q−1)d=y .

Adding, 2a+(p+q−1−1)d=x+y⇒2a+(p+q−1)d=x+y+d .

Therefore,

Sp+q=p+q2[2a+(p+q−1)d]=p+q2[x+y+d]=p+q2[x+y+x−yp−q]