1. Write the monthly savings sequence

For the first three months, he saves ₹$200$ each month.

So the savings are: $200,\ 200,\ 200$

After that, each month the saving increases by ₹$40$ over the immediately previous month.

So from the 4th month onward, the sequence becomes: $240,\ 280,\ 320,\ \dots$

Thus monthly savings are: $200,\ 200,\ 200,\ 240,\ 280,\ 320,\dots$

2. Separate the total saving into two parts

Savings in first 3 months: $3 \times 200 = 600$

Remaining amount to reach ₹$11040$: $11040 - 600 = 10440$

3. Form an AP for savings after the 3rd month

From the 4th month onward, savings form an arithmetic progression with:

• first term: $a = 240$
• common difference: $d = 40$

Suppose there are $n$ terms in this AP. Then: $S_n = \frac{n}{2}[2a + (n-1)d]$

So, $10440 = \frac{n}{2}[2(240) + (n-1)40]$

$10440 = \frac{n}{2}[480 + 40n - 40]$

$10440 = \frac{n}{2}(40n + 440)$

$10440 = 20n(n+11)$

Divide by $20$: $522 = n(n+11)$

$n^2 + 11n - 522 = 0$

4. Solve the quadratic equation

$n^2 + 11n - 522 = 0$

Factorizing: $n^2 + 29n - 18n - 522 = 0$

$n(n+29) - 18(n+29) = 0$

$(n-18)(n+29)=0$

So, $n=18$

(Reject $n=-29$)

5. Find total number of months

These $18$ months are after the first $3$ months.

Hence total months: $18+3=21$

6. Check with options

Option C is: $21 \text{ months}$

So the correct answer is C.