Question

Q.7

Answer 1

for an A.P.,` p^(th) "term" =t_p=20`

`⇒a(p-1)d=20`            ...............(1)

And, `9^(th) "term"=t_q=10`

`⇒a+(q-1)d=10     ............(2)`

subtracting (2) from (1), we get 

`(p-1)d-(q-1)d=10`

`⇒ d(p-1-q+1)=10`

`⇒d=10/(p-q)`

Substituting value of d in (1), we get, 

`a+(p-1)xx10/(p-q)=20`

`⇒ a+(10p-10)/(p-q)=20`

`⇒a=20-(10p-10)/(p-q)`

`⇒a=(20p-20q-10p+10)/(p-q)`

`⇒ a=(10p-20q+10)/(p-q)`

Now, sum of first (p+q) term, 

`s_(p+q)=(p+q)/2[2xx(10p-20q+10)/(p-q)+(p+q-1)xx10/(p-q)]`

`⇒s_(p+q)=(p+q)/2[(20p-40q+20)/(p-q)+(10p+10q-10)/(p-q)]`

`⇒s_(p+q)=(p+q)/2[(20p-40q+20+10p+10q)/(p-q)]`

`⇒s_(p+q)=(p+q)/2[(30p-30q+10)/(p-q)]`

`⇒s_(p+q)=(p+q)/2[(30(p-q))/(p-q)+10/(p-q)]`

`s_(p+q)=(p+q)/2[30+10/(p-q)]`