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Step-by-Step Solution
Step 1: Understand the Problem
A man saves ₹200 each for the first three months. In every subsequent month, his savings increase by ₹40 compared to the previous month’s savings. After some number of months (let that be n), the total savings amount to ₹11040. We need to find n.
Step 2: Break Down the Savings Structure
The total savings for the first three months =
$3 \times 200 = 600.$
From the fourth month onward, the man saves according to an arithmetic progression (AP) whose first term $a = 240$ (since 200 + 40 = 240) and common difference $d = 40.$
If the total number of months is n, then the number of terms in this AP part is $(n - 3)$, because the first three months are already accounted for with a fixed ₹200 each.
Step 3: Set Up the Sum Equation
The total savings can be written as the sum of two parts:
Sum of first 3 months: $600$
Sum of the AP from month 4 to month n:
For an arithmetic progression with $(n-3)$ terms, first term $a = 240$ and common difference $d = 40$, the sum is
$S_{\text{AP}} = \frac{(n - 3)}{2} \bigl[2 \times 240 + (n - 4)\times 40 \bigr].$
So the total savings is:
$S_{\text{total}} = 600 + \frac{(n - 3)}{2} \bigl[2 \times 240 + (n - 4)\times 40 \bigr].$
We know $S_{\text{total}} = 11040.$ Hence,
$600 + \frac{(n - 3)}{2} \bigl[480 + (n - 4)\times 40 \bigr] = 11040.$
Step 4: Simplify the Equation
Subtract 600 from both sides:
$\frac{(n - 3)}{2} \bigl[480 + 40(n - 4) \bigr] = 11040 - 600 = 10440.$
Inside the bracket, simplify $480 + 40(n - 4)$:
$480 + 40n - 160 = 40n + 320.$
So the left side becomes:
$\frac{(n - 3)}{2} [40n + 320] = 10440.$
Multiply both sides by 2:
$(n - 3) [40n + 320] = 20880.$
Factor out 40 from the bracket:
$40 (n - 3) (n + 8) = 20880.$
Divide both sides by 40:
$(n - 3)(n + 8) = \frac{20880}{40} = 522.$
Step 5: Solve the Quadratic Equation
Expand $(n - 3)(n + 8)$:
$n^2 + 8n - 3n - 24 = n^2 + 5n - 24.$
So the equation is:
$n^2 + 5n - 24 = 522.$
Rearrange:
$n^2 + 5n - 546 = 0.$
We can factor this as:
$(n + 26)(n - 21) = 0.$
Thus, the solutions are $n = -26$ or $n = 21.$ Since the number of months cannot be negative, we take
$\boxed{n = 21}.$
Step 6: Conclude
After 21 months, the man's total savings amount to ₹11040. Hence, the required answer is 21 months.
