Question

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

Answer 1

Let the first term of the A.P be $a$ and the common difference be $d$

We have $t_{p+1}=2t_{q+1}$

$\Rightarrow a+(p+1-1)d=2[a+(q+1-1)d]$

$\Rightarrow a+pd=2a+2qd$

$\Rightarrow a=(p-2q)d$ ..........$\left(1\right)$

Now, $t_{3p+1}=a+(3p+1-1)d$

$=(p-2q)d+3pd=4pd-2qd=2(2p-q)d$

$\therefore t_{3p+1}=2(2p-q)d$ ......$\left(2\right)$

and $t_{p+q+1}=a+(p+q+1-1)d=(p-2q)d+(p+q)d$ from $\left(1\right)$

$\Rightarrow t_{p+q+1}=(2p-q)d$ .....$\left(3\right)$

From $\left(2\right)$ and $\left(3\right)$ we have

$t_{3p+1}=2t_{p+q+1}$