Question

Subba Rao started work in $1995$ at an annual salary of $\mathrm{Rs}5000$ and received an increment of $\mathrm{Rs}200$ each year. In which year did his income reach $\mathrm{Rs}7000$ $?$

Answer 1

Finding the year in which income reached $\mathrm{Rs}7000$:

Given that the income of Subba Rao increases every year by $\mathrm{Rs}200$ and his income in $1995$ is $\mathrm{Rs}5000$.

We have to find out the year in which his income reached $\mathrm{Rs}7000$.

After $1995$, the salaries for each year are;

$5000,5200,5400,\dots ...$

Here the common difference between any two consecutive terms is the same, therefore it forms an AP.

In the AP, the first term $=\mathrm{a}=5000$ and common difference $=\mathrm{d}=5200-5000=200$

Let after ${\mathrm{n}}^{\mathrm{th}}$ year, his salary be $\mathrm{Rs}7000$.

Hence, ${\mathrm{a}}_{\mathrm{n}}=7000$

We know that the ${\mathrm{n}}^{\mathrm{th}}$ term of an AP is given by,

${\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}$

Where $\mathrm{n}$ is the number of terms.

By substituting the values in the above equation, we get;

$\begin{array}{rcl}7000& =& 5000+\left(\mathrm{n}-1\right)\times 200\\ & \Rightarrow & 7000=5000+200\mathrm{n}-200\\ & \Rightarrow & 7000=4800+200\mathrm{n}\\ & \Rightarrow & 7000-4800=200\mathrm{n}\\ & \Rightarrow & 2200=200\mathrm{n}\\ & \Rightarrow & 200\mathrm{n}=2200\\ & \Rightarrow & \mathrm{n}=\frac{2200}{200}\\ & \Rightarrow & \mathrm{n}=11\end{array}$

Thus, in the ${11}^{\mathrm{th}}\mathrm{year}$, his income reached $\mathrm{Rs}7000$. This means after $10\mathrm{years}$ of $1995$, that is, $1995+10=2005$

Therefore in $2005$ Subba Rao income reached $\mathrm{Rs}7000$.