Question

The $4^{\text{th}}$ term of an AP is zero. Prove that the ${25}^{\text{th}}$ term of the AP is three times its ${11}^{\text{th}}$ term.

Answer 1

Step 1. Relation between $\text{a}$ and $\text{d}$:

$a_4=a+3d=0$

We know that $a_n=a+(n-1)d$

$a=-3d$

Step 2. Prove ${\mathit{a}}_{\mathbf{25}}\mathbf{=}\mathbf{3}{\mathit{a}}_{\mathbf{11}}$:

$a_{25}=a+24d\dots \dots \dots ..\left(1\right)$

Similarly,

$a_{11}=a+10d\dots \dots \dots ..\left(1\right)$

On substituting the value of $a$ in equations $\left(1\right)$ and $\left(2\right)$ we get,

$\begin{array}{cc}\Rightarrow & a_{11}=7d\\ \Rightarrow & a_{25}=21d\\ \Rightarrow & a_{25}=3\left(7d\right)\\ \Rightarrow & a_{25}=3a_{11}\end{array}$

Hence, it is proved that the ${25}^{th}$ term of the AP is three its ${11}^{\text{th}}$.