A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs. 11040 after.
a = Rs. 200
d = Rs. 40
savings in first two months = Rs. 400
remained savings = 200 + 240 + 280 + ..... upto n terms
\( = \frac{n}{2}\) [400 + (n – 1)40] = 11040 – 400
200n + 20n2 – 20n = 10640
20n2 + 180 n – 10640 = 0
n2 + 9n – 532 = 0
(n + 28) (n – 19) = 0
n = 19
\ no. of months = 19 + 2 = 21 .
This problem describes a man's savings pattern with two distinct phases:
Phase 1 (First 3 months): Constant savings of Rs. 200 each month
Phase 2 (Subsequent months): Savings form an arithmetic progression where each month's saving increases by Rs. 40 more than the previous month
Let n be the total number of months
First 3 months: 200, 200, 200
From month 4 onward: The savings form an AP with first term a and common difference d = 40
The saving in month 4 = 200 + 40 = 240
The saving in month 5 = 240 + 40 = 280, and so on
Total savings = Savings from first 3 months + Savings from remaining (n-3) months
Total savings = 3 × 200 + Sum of AP from month 4 to month n
First term of AP (month 4): a = 240
Number of terms in AP: m = n - 3
Sum of AP =
Total savings =
Total savings = 11040
Subtract 600 from both sides:
Multiply both sides by 2:
Expand:
Divide by 40:
Using quadratic formula:
Taking positive root: n = (-5 + 47)/2 = 42/2 = 21
His total saving will be Rs. 11040 after 21 months
Sum of arithmetic progression:
where a = first term, d = common difference, n = number of terms
Quadratic formula:
for equation