# Ex.5.2 Q19 Arithmetic Progressions Solution - NCERT Maths Class 10

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## Question

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

Video Solution
Arithmetic Progressions
Ex 5.2 | Question 19

## Text Solution

What is Known:?

Annual salary of $$\rm{Rs}\,5000$$ and increment of $$\rm{Rs}\,200$$ each year.

What is Unknown?

The year in which income reached $$\rm{Rs}\,7000$$

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$nth$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

From the Given Data, incomes received by Subba Rao in the years $$1995,\,1996,\,1997\,\dots$$ are $$5000,\,5200, \,5400,\,\dots 7,000$$

From Observation,

\begin{align}a&= 5000\\d &= 200\end{align}

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

Hence $${a_n} = 7000$$

We know that the $$n^\rm{th}$$ term of an A.P. Series,

${a_n} = a{\rm{ }} + \left( {n - 1} \right)d$

By Substituting above values,

\begin{align}7000 &= 5000 + (n - 1)200\\200(n - 1) &= 7000 - 5000\\n - 1 &= \frac{{2000}}{{200}}\\ n &= 10 + 1\\n &= 11\end{align}

Therefore, in $$11^\rm{th}$$ year, his income reached $$\rm{Rs}\,7000.$$ Which means after $$10$$ years of $$1995$$ i.e. $$1995 + 10 \Rightarrow 2005$$

In $$2005$$ his income reached $$\rm{Rs}\,7000$$.