Question

A man saves Rs 32,000 during first year, Rs 36,000 in the next year and Rs 40,000 in the third year.
If he continues his savings in this sequence, in how many years will he saves Rs 2,00,000 ?

Answer 1

Answer:

In the 43rd year the savings will become 2,00,000 rupees

Step-by-step Explanation :

Given : A man saves :

1) 32000 Rs ---------- first year

2) 36000 Rs ----------- second year

3) 40000 Rs ------------- third year

To find : In how many years will he saves Rs 2,00,000 = ?

The savings are increased in arithmetic sequence.

The nth term of arithmetic = a1 + (n-1) × d

sequence

Where : a1 = first term of A.P.

n = nth term of A.P.

d = Common difference

From the given data we can say that,

a1 = 32000 & d = 4000

Substituting the given value in above formula we get,

200000 = 32000 + (n-1) × 4000

200000 - 32000 = (n-1) × 4000

168000 = (n-1) × 4000

168000/4000 = (n-1)

n-1 = 42

n = 42+1 = 43

Therefore In the 43rd year the savings will become 2,00,000 rupees