Ex 5.2 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,
Subba Rao's annual salary in year 1995 was ₹5000.
➙ Since he received an increment of ₹200 each year,
∴ Salary in 1996 (2nd year)
= 5000 + 200
= 5200
∴ Salary in 1997 (3rd year)
= 5200 + 200
= 5400
∴ Annual salary received by Subba Rao in year 1995, 1996, 1997 was...
5000, 5200, 5400...
*Clearly, this forms an AP with
a = 5000 d = 200
Let,
Subba Rao's annual salary reached to ₹7000 in nth year,
∴ aₙ = 7000
*Using formula, aₙ = a + (n - 1)d
∴ a + (n - 1)d = 7000
∴ 5000 + (n - 1)200 = 7000
∴ 5000 + 200n - 200 = 7000
∴ 200n = 7000 - 5000 + 200
∴ 200n = 2200
∴ n =
2200
/
200
∴ n = 11
Therefore, Subba Rao received ₹7000 in 11th year's annual salary.
*Finding 11th year
1st year = 1995
2nd year = 1996 { 1995 + 1 }
3rd year = 1997 { 1995 + 2 }
.
.
11th year = 1995 + 10
= 2005
Hence,
Subba Rao's income reached ₹7000 in the year 2005.
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Try This..
✒ An employee’s salary in the first year is ₹8000, and it increases by ₹500 every year. After how many years will it become ₹13000?
✒ The first term of an AP is 200 and the common difference is ₹100. Find the year when it reaches ₹1000.
✒ A person starts saving ₹100 per month, increasing it by ₹50 each month. In which month will the saving be ₹1000?
❌ Common Mistakes
Forgetting that the first year is included:▸ The nth term means "after n terms," but we need to account for the starting year when converting to calendar years.Misplacing the formula:▸ Always start with aₙ = a + (n - 1)d, not aₙ = a + nd.Arithmetic errors:▸ Small calculation mistakes can lead to wrong answers.▸ Double-check your subtraction and division steps.
Queries Solved:-
Class 10 Ex 5.2
Ex 5.2 Q19 Class 10
Class 10 Ex 5.2 Q19
Class 10 Chap 5 Ex 5.2 Q19
Class 10 Arithmetic Progression
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