Question

Given $d=5$ and $S_9=75$. Find the value of $a$ and $a_9$

Answer 1

Step 1: Find the value of $a$

We know that the sum of the first '$n$' terms of an arithmetic progression is given by

${\mathit{S}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{2}}\left[2a+(n-1)d\right]$;

substituting $n=9$ ; $S_9=75$ and $d=5$ in the standard result, we get

$$\begin{gathered} 75=\frac{9}{2}\left[2a+(9-1)5\right] \\ \Rightarrow 150=9(2a+40) \\ \Rightarrow 150=18a+360 \\ \Rightarrow a=\frac{-210}{18} \\ \Rightarrow a=\frac{-35}{3} \end{gathered}$$

Step 2: Find the value of $a_9$

We know that the $n^{th}$ term of an arithmetic progression is given by

${\mathit{a}}_{\mathbf{n}}\mathbf{}\mathbf{=}\mathit{a}\mathbf{+}\mathbf{(}\mathit{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{d}$

substituting $a=\frac{-35}{3}$ ; $n=9$ and $d=5$ in the standard result we get,

$$\begin{gathered} a_9=\frac{-35}{3}+(9-1)5 \\ \Rightarrow a_9=\frac{-35}{3}+40 \\ \Rightarrow a_9=\frac{85}{3} \end{gathered}$$

Hence value of $\mathit{a}\mathbf{=}\frac{\mathbf{-}\mathbf{35}}{\mathbf{3}}$ and ${\mathit{a}}_{\mathbf{9}}\mathbf{}\mathbf{=}\frac{\mathbf{85}}{\mathbf{3}}$.