Question

Subba rao started the work in 1995 at an annual salary Rs. 5000 and received an increment of Rs. 200 every year. In which year did his income reached Rs. 7000?
(a) 2005
(b) 2002
(c) 2004
(d) 2007

Answer 1

Hint: We solve this problem by using the arithmetic progression.
The general representation of A.P is given as
\[a,a+d,a+2d,.....\]
Where, \[a\] is the first term and \[d\] is the common term.
We have the formula for \[{{n}^{th}}\] term of an A.P as
\[{{T}_{n}}=a+\left( n-1 \right)d\]
By using the above formula we find the number of years it takes to get Rs. 7000 so that we get the year by adding a number of years to 1995.

Complete step by step answer:
We are given that the initial salary in the year 1995 as Rs. 5000
Let us assume that the first salary as
\[\Rightarrow a=5000\]
We are given that there is an increment of Rs 200 every year.
Here, we can see that the salary increment gives the form of A.P
We know that the general representation of A.P is given as
\[a,a+d,a+2d,.....\]
Where \[a\] is the first term and \[d\] is the common term.
Let us assume that the given increment as
\[\Rightarrow d=200\]
We are asked to find the year at which the salary becomes Rs. 7000
Let us assume that the final amount as
\[\Rightarrow {{T}_{n}}=7000\]
We know that the formula for \[{{n}^{th}}\] term of an A.P as
\[{{T}_{n}}=a+\left( n-1 \right)d\]
By using the above formula to given information then we get
\[\begin{align}
  & \Rightarrow 7000=5000+\left( n-1 \right)200 \\
 & \Rightarrow 2000=\left( n-1 \right)200 \\
 & \Rightarrow n-1=10 \\
 & \Rightarrow n=11 \\
\end{align}\]
Therefore, we can say that at the end of \[{{11}^{th}}\] year form 1995 the salary becomes Rs. 7000
We know that the year after \[r\] years from \[n\] is given as \[n+r-1\]
Let us add 11 to 1996 by using the above condition then we get
\[\Rightarrow 1995+11-1=2005\]
Therefore, we can conclude that in the year 2005 the salary becomes Rs. 7000
So, option ( a) is correct answer.

Note:
Students may do mistakes in taking the year after the 11 years.
We consider that the year 1995 as the first year in that case we need to subtract 1 after adding 11 to 1995 then we get
\[\Rightarrow 1995+11-1=2005\]
But students may directly add 11 to 1995 and gives the answer. This gives the wrong answer because in this case, we consider the next year that is 1996 as the first year of the series.