Question

Find the ${31}^{\mathrm{st}}$ term of the AP whose ${11}^{\mathrm{th}}$ term is $38$ and the ${16}^{\mathrm{th}}$ term is $73$.

Answer 1

Step 1: Form equations from the given information

Given, ${11}^{\mathrm{th}}$ term, that is $a_{11}=38$

and, ${16}^{\mathrm{th}}$ term, that is $a_{16}=73$

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

So, the required equations are,

$$\begin{gathered} \Rightarrow 38=a+10d...\left(1\right) \\ \Rightarrow 73=a+15d...\left(2\right) \end{gathered}$$

Solve the two equations by replacing $a=38-10d$ in equation $\left(2\right)$,

$$\begin{gathered} \Rightarrow 73=a+15d \\ \Rightarrow 73=38-10d+15d \\ \Rightarrow 5d=73-38 \\ \Rightarrow d=\frac{35}{5} \\ \Rightarrow d=7 \end{gathered}$$

Now, replace $d=7$ in equation $\left(1\right)$, we get

$$\begin{gathered} \Rightarrow 38=a+10d \\ \Rightarrow 38=a+(10\times 7) \\ \Rightarrow a=38-70 \\ \Rightarrow a=-32 \end{gathered}$$

So, the first term is $-32$

Step 2: Evaluate the required term

We have $a=-32,d=7$ and $n=31$

$$\begin{gathered} \Rightarrow a_{31}=a+(31-1)d \\ \Rightarrow a_{31}=-32+(30\times 7) \\ \Rightarrow a_{31}=-32+210 \\ \Rightarrow a_{31}=178 \end{gathered}$$

Therefore, the ${31}^{\mathrm{st}}$ term of the AP is $178$.