Ramesh saves money each month in an arithmetic progression. He saves Rs 1500 in the first month and each subsequent month he saves Rs 250 more than the previous month. After how many months will his total savings become at least Rs 2,00,000? Also state the total amount saved at the end of that month.

hard 5 marks Class 10 · Mathematics · Arithmetic Progressions

Answer

35 months, Rs 2,01,250

The correct answer is 35 months, Rs 2,01,250 — Thus after 35 months the total saved is Rs 2,01,250, which is the first amount $g e 2, 00, 000$ .

Step-by-step solution

Identify the AP parameters: first term $a = 1500$ , common difference $d = 250$ .
The sum after $n$ months is $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ .
Set the inequality $S_{n} \geq 200000$ .
Substitute $a$ and $d$ : $\frac{n}{2} [3000 + (n - 1) 250] \geq 200000$ .
Simplify the bracket: $3000 + 250 n - 250 = 250 n + 2750$ .
Thus $S_{n} = \frac{n ( 250 n + 2750 )}{2} \geq 200000$ .
Multiply by 2: $250 n^{2} + 2750 n \geq 400000$ .
Divide by 250: $n^{2} + 11 n - 1600 \geq 0$ .
Solve the quadratic $n^{2} + 11 n - 1600 = 0$ using the formula: $n = \frac{- 11 \pm 1 1 ^{2} + 4 \times 1600}{2} = \frac{- 11 \pm 6521}{2}$ .
$6521 \approx 80.78$ . Take the positive root: $n = \frac{- 11 + 80.78}{2} \approx 34.89$ .
Since $n$ must be an integer month, the smallest integer satisfying the inequality is $n = 35$ .
Check $S_{35}$ : $S_{35} = \frac{35}{2} [3000 + 34 \times 250] = \frac{35}{2} [3000 + 8500] = \frac{35}{2} \times 11500 = 35 \times 5750 = 201250$ .
Thus after 35 months the total saved is Rs 2,01,250, which is the first amount $g e 2, 00, 000$ .

Common mistake: Solving the quadratic incorrectly or forgetting to round up to the next whole month.

Concept tested: application of sum of AP to real‑world problem

Published 23 August 2026 · Last updated 16 September 2026 · How this solution was prepared and checked