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

Chapter number- 9

Exercise 9.6

In this question we have been given that a man saved Rs 32 during the first year, Rs 36 during the second year and in this way he increases his savings by Rs 4 per year.

Now we have to find In what time his savings would be Rs 200 by using the arithmetic progression based concept.

CBSE DHANPAT RAI publications

Class:- 10th

Solutions of CBSE Mathematics

Question 64

So, the first term(a) = 32 and common difference(d) = 4.

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

Sn = n[2a +(n â€“ 1)d] / 2

Here, Sn = 200

So,

=> n[2(32) + (n â€“ 1)4] / 2 = 200

=> n[64 + 4n â€“ 4] / 2 = 200

=> n[60 + 4n] / 2 = 200

=> 2n2 + 30n â€“ 200 = 0

=> n2 + 15n â€“ 100 = 0

=> n2 + 20n â€“ 5n â€“ 100 = 0

=> n(n + 20) â€“ 5(n + 20) = 0

=> (n â€“ 5) (n + 20) = 0

=> n = 5 or n = â€“20

Ignoring n = â€“20 as number of terms cannot be negative. So, we get n = 5.

Therefore, in 5 years his savings would be Rs 200.