Question

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

Answer 1

Hint: In this particular question use the concept of Arithmetic progression, the last term of A.P is given as ${a_n} = a + \left( {n - 1} \right)d$, where a is the first term, d is the common difference, n is the number of terms in the A.P and ${a_n}$ is the last term, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given data:
Subba Rao annual salary in 1995 is Rs. 5000.
Now he receives an increment of Rs. 200 each year.
So next year his annual salary is 5200, next year, 5400 and so on.
Now we have to find out in which year his salary became Rs. 7000.
Now assume this as an arithmetic series whose last term is 7000 so we have,
5000, 5200, 5400, ........... 7000
So the first term of the series is, a = 5000.
Common difference, d = (5200 – 5000) = (5400 – 5200) = 200
And last term is, ${a_n}$ = 7000.
Let the number of terms in the series are n.
Now as we know that the last term of A.P is given as ${a_n} = a + \left( {n - 1} \right)d$, where a is the first term, d is the common difference, n is the number of terms in the A.P and ${a_n}$ is the last term.
$ \Rightarrow {a_n} = a + \left( {n - 1} \right)d$
Now substitute the values we have,
$ \Rightarrow 7000 = 5000 + \left( {n - 1} \right)200$
$ \Rightarrow 7000 - 5000 = \left( {n - 1} \right)200$
$ \Rightarrow \left( {n - 1} \right) = \dfrac{{2000}}{{200}} = 10$
$ \Rightarrow n = 10 + 1 = 11$
So the total terms in the series is 11.
So if starting year is 1995, so the year in which his salary become Rs. 7000 is,
Year 1995 + (11 – 1) years = 2005 year, (minus 1 is for current year)
So his salary became Rs. 7000 in 2005.
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that reduce the problem into a standard series (i.e. G.P, A.P, or a H.P) according to the given data in the question, then calculate the necessary parameters of this series to calculate the number of terms in the series as above, then add one less than number of terms in the current year we will get the required year in which his salary become Rs. 7000.