This is the basic and conceptual question from Chapter name- Arithmetic Progression

Chapter number- 9

Exercise 9.6

In this question its is given that Resham wanted to save at least Rs. 6500 for sending her daughter to school next year (after 12 months).

She saved Rs. 450 in the first month and raised her saying by Rs.20 every next month.

Now we have to find how much will she be able to save in next 12 months

Also we have to find out if she will be able to send her daughter to the school next year..

CBSE DHANPAT RAI publications

Class:- 10th

Solutions of CBSE Mathematics

Question 60

Given that Resham has saved in following sequence in a year (every month record): 450, 470, 490,. . . .

This sequence is an A.P. with first term(a) = 450 and common difference(d) = 20.

Now by using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

So,

Now, sum of the savings after 12 months would be S12 as value of n in this case is 12.

S12 = 12[2(450) + (12 – 1)20] / 2

= 6[900 + 220]

= 6[1120]

= 6720

So, Resham would save Rs 6720 by the end of 12 months which is greater than Rs 6500,

she would be able to send her daughter to the school next year.