Question

If $10$ times the ${10}^{\mathrm{th}}$ term of an AP is equal to $15$ times the ${15}^{\mathrm{th}}$ term. Show that ${25}^{\mathrm{th}}$ term of the AP is zero.

Answer 1

Step 1: Find the terms using the AP formula.

The formula for finding the $n^{\mathrm{th}}$ term of an AP is $a_n=a+(n-1)d$,

Where $a$is the first term, $n$ is the number of terms and $d$ is the difference between the terms.

$\begin{array}{rcl}{a}_{10}& =& a+(10-1)d\\ \therefore a_{10}& =& a+9d\\ a_{15}& =& a+(15-1)d\\ \therefore a_{15}& =& a+14d\end{array}$

Step 2: Solve for the value of $a$ with respect to the condition.

According to the condition,

$10\left(a_{10}\right)=15\left(a_{15}\right)$

Substitute the respective values and solve.

$\begin{array}{rcl}& & \begin{array}{cc}& 10(a+9d)=15(a+14d)\\ \Rightarrow & 10a+90d=15a+210d\\ \Rightarrow & -5a+90d=210d\\ \Rightarrow & -5a=120d\\ \Rightarrow & a=-24d\end{array}\end{array}$

Step 3: Put the value of $a$ in the ${25}^{\mathrm{th}}$ term.

$\begin{array}{rcl}{a}_{25}& =& a+(25-1)d\\ & \Rightarrow & a_{25}=-24d+24d\\ & \Rightarrow & a_{25}=0\end{array}$

Hence, it is proved that, the ${25}^{\mathrm{th}}$ term of an AP is zero, if $10$ times its ${10}^{\mathrm{th}}$ term is equal to $15$ times its ${15}^{\mathrm{th}}$ term.