Question

The ${11}^{\mathrm{th}}$ term of an AP is $80$ and the ${16}^{\mathrm{th}}$ term is $110$. Find the ${31}^{\mathrm{st}}$ term.

Answer 1

Step 1. Find the general term for ${11}^{\mathrm{th}}$ and ${16}^{\mathrm{th}}$ term:

We know that the nth term of AP is ${\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}$.

Therefore,

$$\begin{gathered} \Rightarrow a_{11}=\mathrm{a}+(11-1)\mathrm{d} \\ \Rightarrow 80=\mathrm{a}+10\mathrm{d}......\left(1\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow a_{16}=\mathrm{a}+(16-1)\mathrm{d} \\ \Rightarrow 110=\mathrm{a}+15\mathrm{d}......\left(2\right) \end{gathered}$$

Step 2. Subtract the equation (1) and (2):

We get,

$$\begin{gathered} \Rightarrow \mathrm{a}+15\mathrm{d}-\mathrm{a}-10\mathrm{d}=110-80 \\ \Rightarrow 5\mathrm{d}=30 \\ \Rightarrow \mathrm{d}=6 \\ \Rightarrow \mathrm{a}+10\times 6=80\left[\because \text{Eqn}\left(2\right)\right] \\ \Rightarrow \mathrm{a}=20 \end{gathered}$$

On substituting in $a_n$

$\begin{array}{rcl}& \Rightarrow & a_{31}=a+(31-1)d\end{array}$

$\begin{array}{rcl}& \Rightarrow & a_{31}=20+180\\ & =& 200\end{array}$

Hence, the ${31}^{\mathrm{st}}$ term for the A.P is $200$