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

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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 \(\rm{Rs}\,7000\)?

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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 \(\rm{Rs}\,7000\)


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


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 \(\rm{Rs}\,7000\).