Question

Given that the $(p+1{)}^{th}$ term of an A.P is twice the $(q+1{)}^{th}$ term, prove that the $(3p+1{)}^{th}$ term is twice the $(p+q+1{)}^{th}$ term.

Answer 1

Let the first term be 'a'

the common difference is 'd'

Now$(AP{)}^{th}$term is

$=a+(n-1)d$

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

$=a+pd------\left(1\right)$

$(p+1{)}^{th}=2(q+1{)}^{th}$ term is

$=a+(q+1-1).d$

$=a+qd------\left(2\right)$

According to question

$a+pd=2(a+qd)$

$a+pd=2a+2qd$

$a=d[p-2q]-----\left(3\right)$

Now to prove

$(3p+1{)}^{th}term=2\times [p+q+1{]}^{th}term$

$\Rightarrow a+(3p+1-1)d.=2\times [a+[p+q+1-1\left]d\right]$

$\Rightarrow a+3pd=2[a+(p+q\left)d\right]$

$\Rightarrow a=3pd-2(p+q).d$

$\Rightarrow a=3pd-2pd-2qd$

$\Rightarrow a=pd-2qd$

$\Rightarrow a=d[p-2q]----\left(4\right)$

form (3)put the value of $a$

$\Rightarrow d[p-2q]=d[p-2q]$

$\therefore$ $L.H.S=R.H.S$