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Q19:
Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?

Solution :

Given:

The starting annual salary of Subba Rao in the year 1995 is $a = 5000$.

The annual increment is $d = 200$.

The target annual salary is $a_n = 7000$.

To Find:

The year in which his annual income reaches ₹ 7000.

Step 1: Identifying the Arithmetic Progression (AP)

Since the salary increases by a fixed amount every year, the sequence of annual salaries forms an Arithmetic Progression.

The general form of an AP is given by: $a, a+d, a+2d, \dots, a+(n-1)d$.

Here, the first term $a = 5000$ and the common difference $d = 200$.

Step 2: Applying the Formula for the $n^{th}$ term

The formula for the $n^{th}$ term of an Arithmetic Progression is:

$a_n = a + (n - 1)d$

Substituting the given values into the formula:

$7000 = 5000 + (n - 1)200$

Step 3: Solving for $n$

Subtract 5000 from both sides of the equation:

$7000 - 5000 = (n - 1)200$

$2000 = (n - 1)200$

Divide both sides by 200:

$\frac{2000}{200} = n - 1$

$10 = n - 1$

Add 1 to both sides:

$n = 10 + 1$

$n = 11$

Step 4: Determining the Year

The salary reached ₹ 7000 in the $11^{th}$ year of his service.

Since he started in 1995, the year is calculated as:

Year = (Starting Year) + $(n - 1)$

Year = $1995 + (11 - 1)$

Year = $1995 + 10$

Year = $2005$

Final Answer: Subba Rao's income reached ₹ 7000 in the year 2005.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.2


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


Chapters in CBSE - Class 10 Mathematics


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