Question

If $9$ times the $9^{th}$ term of an AP is equal to $13$times the${13}^{th}$ term, then the ${22}^{nd}$ term of the AP is

• A. $0$
• B. $22$
• C. $220$
• D. $198$

Answer 1

The correct option is A

$0$

Explanation for the correct option

Step 1: Solve for the first term

Given that $9$ times the $9^{th}$ term of an AP is equal to $13$times the${13}^{th}$ term

$\begin{array}{crclc}\Rightarrow & 9\times a_9& =& 13\times a_{13}& \\ \Rightarrow & 9[a+(9–1\left)d\right]& =& 13[a+(13–1\left)d\right]& \left[\because a_n=a+\left(n-1\right)d\right]\\ \Rightarrow & 9[a+8d]& =& 13[a+12d]& \\ \Rightarrow & 9a+72d& =& 13a+156d& \\ \Rightarrow & 13a-9a& =& 72d-156d& \\ \Rightarrow & 4a& =& -84d& \\ \Rightarrow & a& =& -21d& \end{array}$

Step 2: Solve for the ${22}^{nd}$ term

$\begin{array}{ccclc}\therefore & a_{22}& =& a+\left(22-1\right)d& \\ & & =& a+21d& \\ & & =& -21d+21d& \left[\because a=-21d\right]\\ \Rightarrow & a_{22}& =& 0& \end{array}$

Hence the correct option is option(A) i.e.$0$