Arithmetic progressions - Problems based on arithmetic progressions for class 10 maths. Lets tute is an online learning centre. We provide quality education for all learners and 24/7 academic guidance through E-tutoring. Our Mission- Our aspiration is to be a renowned unpaid school on Web-World. Contact Us - Website - www.letstute.com YouTube - www.youtube.com/letstute

1. Arithmetic Progressions
Problems based on
Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com

2. Problems based on
Arithmetic Progressions
Q) 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?
Given: Annual salary = Rs 5000
Increment = Rs 200
To find: Year in which the income reached
Rs 7000
Chapter : Arithmetic Progressions Website: www.letstute.com

3. Problems based on
Arithmetic Progressions
Solution: Annual salary received by Subba Rao in 1995, 1996,
1997,… is Rs 5000, Rs 5200, Rs 5400,…
Sequence of salaries (in Rs) 5000, 5200, 5400, --- forms an
arithmetic progression with first term a = 5000 and common
difference d = 200.
Chapter : Arithmetic Progressions Website: www.letstute.com

4. Problems based on
Arithmetic Progressions
 a + (n - 1)d = 7000
 5000 + (n - 1) (200) = 7000
 200n - 200 = 7000 - 5000
 200n = 2200
 n = 11
Thus, in the 11th year Subba Rao’s income
reached Rs. 7000.
Chapter : Arithmetic Progressions Website: www.letstute.com

5. Problems based on
Arithmetic Progressions
Q) Ramkali saved Rs 5 in the first week of a year and then
increased her weekly savings by Rs 1.75. If in the nth week,
her weekly saving become Rs. 20.75, find n.
Given: Savings in the 1st week of the year = Rs 5
Increase in weekly savings = Rs 1.75
To find: n
Chapter : Arithmetic Progressions Website: www.letstute.com

6. Problems based on
Arithmetic Progressions
Solution: Ramkali’s weekly savings are Rs. 5, Rs. 6.75, Rs.8.50…..
Her weekly savings in nth week = 20.75
nth term = 20.75
 a + (n – 1)d = 20.75
 5 + (n – 1)(1.75) = 20.75
Chapter : Arithmetic Progressions Website: www.letstute.com

7.  (1.75n – 1.75) = 20.75 - 5
 1.75n = 15.75 + 1.75
 1.75n = 17.50

n = 10
Hence, n = 10
Problems based on
Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com

8. Problems based on
Arithmetic Progressions
Q) Compute the sum of first 30 terms of the AP 3, 8, 13, 18,...
Solution:
First term = a = 3, common difference = d = 5, n = 30
S30 = 3 0 [(2)(3) + (30 – 1)(5)]
2

Sn = n [2a + (n – 1)d]
2
 S30 = 15 [6+ (29)(5)]
 S30 = 15 [6+ 145]
 S30 = 15 [151]
S30 = 2265
Hence, the sum of first 30 terms of the given AP is 2265
Chapter : Arithmetic Progressions Website: www.letstute.com

9. Problems based on
Arithmetic Progressions
Q)Find the sum of the first 200 natural numbers.
Solution:
Sum of 200 natural numbers
= 1 + 2 + 3 + ….+ 199 + 200.
Here, a = 1, d = 2 – 1 = 1, n = 200
Sn = n [2a + (n – 1)d]
2
200
2
 S200 = [2(1) + (200 – 1)(1)]
 S200 = 1 0 0 (2 + 199)
 S200 = 1 0 0 (201) = 20100
Hence, the sum of first 200 natural numbers is 20100
Chapter : Arithmetic Progressions Website: www.letstute.com