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