Question

5. If $7$ times of $7^{th}$ term of an AP is equal to $11$times of ${11}^{th}$term, show that the ${18}^{th}$term of the AP is zero.

Answer 1

Step 1. Note the given data:

The $7^{th}$term of the AP is $t_7$ and ${11}^{th}$term of the AP is $t_{11}$.

Given $7t_7=11t_{11}$….(i)

Let $a$ be the first term of the AP and $d$ is the common ratio of the AP.

Step 2. Compute $t_7$ in terms of $a$ and $d$:

The formula of $n^{th}$term of an AP is $t_n=a+\left(n-1\right)d$.

For $t_7$, $n=7$

$t_7=a+(7-1)d$

$=a+6d$

Step 3. Compute $t_{11}$in terms of $a$ and $d$:

The formula of $n^{th}$term of an AP is $t_n=a+\left(n-1\right)d$.

For $t_{11}$, $n=11$

$t_{11}=a+(11-1)d$

$=a+10d$

Step 4. Plugging the values of $t_7$ and $t_{11}$ in equation (i)

We have $7t_7=11t_{11}$

$$\begin{gathered} 7\left(a+6d\right)=11\left(a+10d\right) \\ 7a+42d=11a+110d \end{gathered}$$

$11a-7a+110d-42d=0$

$4a+68d=0$

$\Rightarrow a=-17d$

Step 5. Finding the ${18}^{th}$term of the AP

The formula of $n^{th}$term of an AP is $t_n=a+\left(n-1\right)d$.

For $t_{18}$, $n=18$

$t_{18}=a+(18-1)d$

$=a+17d$

Step 6. Plugging $a=-17d$ in the above equation

$$\begin{gathered} t_{18}=a+17d \\ =-17d+17d \\ =0 \end{gathered}$$

Hence, the ${18}^{th}$term of the AP is zero.