Arithmetic progressions - Problems based on arithmetic progressions Part - 5 for class 10 maths

Chapter : Arithmetic Progressions Website: www.letstute.com

Arithmetic Progressions

Problems based on Arithmetic Progressions

Given: Annual salary = Rs 5000 Increment = Rs 200

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?



To find: Year in which the income reached Rs 7000

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.



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.



a + (n - 1)d = 7000

Given: Savings in the 1st week of the year = Rs 5 Increase in weekly savings = Rs 1.75

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.



To find: n

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



(1.75n – 1.75) = 20.75 - 5

1.75n = 15.75 + 1.75

1.75n = 17.50

n = 10

Hence, n = 10



Solution:

Q) Compute the sum of first 30 terms of the AP 3, 8, 13, 18,...

First term = a = 3, common difference = d = 5, n = 30

S30 = [(2)(3) + (30 – 1)(5)]30 2

Sn = [2a + (n – 1)d]n2

S30 = 15 [6+ (29)(5)]

S30 = 15 [6+ 145]

S30 = 15 [151]

Hence, the sum of first 30 terms of the given AP is 2265



S30 = 2265

Solution:

Q)Find the sum of the first 200 natural numbers.

Sum of 200 natural numbers = 1 + 2 + 3 + ….+ 199 + 200.

Sn = [2a + (n – 1)d] n 2

Here, a = 1, d = 2 – 1 = 1, n = 200

Hence, the sum of first 200 natural numbers is 20100

S200 = [2(1) + (200 – 1)(1)] 200 2

S200 = (2 + 199) 100

S200 = (201) = 20100 100



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Arithmetic progressions - Problems based on arithmetic progressions Part - 5 for class 10 maths