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 – 1)d] / 2
Here, Sn = 200
So,
=> n[2(32) + (n – 1)4] / 2 = 200
=> n[64 + 4n – 4] / 2 = 200
=> n[60 + 4n] / 2 = 200
=> 2n2 + 30n – 200 = 0
=> n2 + 15n – 100 = 0
=> n2 + 20n – 5n – 100 = 0
=> n(n + 20) – 5(n + 20) = 0
=> (n – 5) (n + 20) = 0
=> n = 5 or n = –20
Ignoring n = –20 as number of terms cannot be negative. So, we get n = 5.
Therefore, in 5 years his savings would be Rs 200.