One of the most important and exam oriented question from Chapter name- Arithmetic Progression

Class 10th

Chapter number- 9

Exercise :- 9.6

This type of question has been asked in previous years exams.

In this question we have been given that a man saved Rs 16500 in ten years.

In each year after the first year, he saved Rs 100 more than he did in the preceding year.

Now we have to find how much did he save in the first year.

We have to solve this question by using the arithmetic progression based concept.

CBSE DHANPAT RAI publication

CBSE Mathematics Class 10th

Question 63

the common difference(d) = 100 and sum of savings done in 10 years(S10) = 16500

We have to find the savings in first year, i.e., first term(a).

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

Sn = n[2a +(n – 1)d] / 2

So,

=> S10 = 16500

=> 10[2a + (10 – 1)100] / 2 = 16500

=> 5[2a + 900] = 16500

=> 10a = 16500 – 4500

=> 10a = 12000

=> a = 1200

Hence, value of savings in first year is 1200.