A man saves Rs 200 in each of the first three months of his service. In each of the subsequent month

Q: Why can't we just use the sum of an A.P. formula for the entire duration of $n$ months?

That is a classic trap! The problem states that the savings are constant for the first three months and only start increasing from the fourth month. An A.P. formula assumes the pattern applies to every single term from the start. Since the first three months don't follow the $d=40$ increase, we must treat them separately as a constant sum and then apply the A.P. formula only to the remaining $n-3$ months.

Q: Why do we reject the negative root when solving the quadratic equation?

In mathematics, quadratic equations often yield two solutions. However, in physics and real-world problems like this, the variable $n$ represents time (number of months). Time cannot be negative. Therefore, we discard the negative root as it is physically impossible in this context.

Q: How do I determine the first term 'a' for the A.P. phase?

The problem says the savings increase by Rs $40$ *more than the saving of the immediately previous month* starting from the 4th month. The 3rd month's saving was Rs $200$. So, the 4th month's saving is $200 + 40 = 240$. This $240$ is the first term of your A.P. phase, not $200$.

Question

A man saves Rs $200$ in each of the first three 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

Options:

(A) 19 months

(B) 20 months

(C) 21 months

(D) 18 months

Sigma Percentile · Accepted Answer

### Analyzing the Savings Pattern

The problem requires us to model a savings trajectory that transitions from a constant phase to a growth phase. We are tasked with finding the total time $n$ required to reach a target of Rs $11040$.

### Phase 1: The Foundation

For the first three months, the man saves a constant amount of Rs $200$ per month. The total savings accumulated during this initial period is:

$$3 	imes 200 = 600$$

Given the total target of Rs $11040$, we subtract the baseline savings to determine the remaining amount that must be covered by the growth phase:

$$11040 - 600 = 10440$$

### Phase 2: The Growth Engine

Starting from the 4th month, the savings increase by Rs $40$ each month. Since the 3rd month's savings were Rs $200$, the 4th month's savings are $200 + 40 = 240$.

This sequence forms an Arithmetic Progression (A.P.) where the first term $a = 240$ and the common difference $d = 40$. If the total duration is $n$ months, the number of months in this A.P. phase is $m = n - 3$.

### The Master Equation

The sum of an A.P. is given by the formula $S_m = \frac{m}{2}[2a + (m-1)d]$. Substituting our known values into this equation, we get:

$$10440 = \frac{n-3}{2}[2(240) + (n-3-1)40]$$

Simplifying the expression inside the brackets:

$$10440 = \frac{n-3}{2}[480 + (n-4)40]$$

Dividing the terms inside the bracket by $2$:

$$(n-3)[240 + (n-4)20] = 10440$$

Expanding the terms further:

$$(n-3)[240 + 20n - 80] = 10440$$
$$(n-3)[20n + 160] = 10440$$

Factoring out $20$ and dividing both sides by $20$:

$$(n-3)(n+8) = 522$$
$$n^2 + 5n - 24 = 522$$
$$n^2 + 5n - 546 = 0$$

### Final Calculation

We solve the quadratic equation $n^2 + 5n - 546 = 0$ by factoring. We look for two numbers that multiply to $-546$ and add to $5$, which are $26$ and $-21$:

$$(n+26)(n-21) = 0$$

This yields two potential solutions: $n = -26$ and $n = 21$. Since time cannot be negative, we discard the negative root.

The total time required to reach the target is **$21$ months**.

Select Answer:

Visualized Solution

Visualizing the Savings Pattern

Phase 1: Constant Savings

Calculating the Remaining Target

Phase 2: Identifying the A.P.

Defining the Number of Terms

Setting up the Sum Equation

Substituting $m = n - 3$

Simplifying the AP Sum Expression

Reducing the Equation

Factoring out 20

Forming the Quadratic Equation

Solving the Quadratic Equation

Finding the Valid Value of $n$

Final Conclusion

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Savings Pattern

Phase 1

Phase 2

The Master Equation

Final Calculation

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