For the total savings S_n to exceed ₹20,000, we set S_n > 20000. Simplifying the expression inside gives (n)/(2)[1000 + 100n - 100] = (n)/(2)[100n + 900] = 50n(n + 9) > 20000, which reduces to n(n + 9) > 400, or n² + 9n - 400 > 0.

For the total savings S_n to exceed ₹20,000, we set S_n > 20000. Simplifying the expression inside gives (n)/(2)[1000 + 100n - 100] = (n)/(2)[100n + 900] = 50n(n + 9) > 20000, which reduces to n(n + 9) > 400, or n² + 9n - 400 > 0.

Solving the equation n² + 9n - 400 = 0 by factoring, we get (n + 25)(n - 16) = 0. Since n must be a positive integer, n = 16. At n = 16, the total savings is exactly 16 × 25 × 50 = 20000 rupees. Therefore, for the total savings to strictly exceed ₹20,000, the minimum number of months required is 17.

Solving the equation n² + 9n - 400 = 0 by factoring, we get (n + 25)(n - 16) = 0. Since n must be a positive integer, n = 16. At n = 16, the total savings is exactly 16 × 25 × 50 = 20000 rupees. Therefore, for the total savings to strictly exceed ₹20,000, the minimum number of months required is 17.

a man saves 500 rupees in the first month and English his monthly savings by 100 every month fine is savings in the 12th month the total amount saved in 12 months the minimum number of months required for his total savings to exceed 20000 rupees

Answer: (i) Savings in the 12th month = ₹1,600 (ii) Total amount saved in 12 months = ₹12,600 (iii) Minimum number of months to exceed ₹20,000 = 17 months

Step-by-step solution

Step 1: Identify the arithmetic progression

The savings in consecutive months are ₹500, ₹600, ₹700, and so on. Since the monthly savings increase by a fixed amount of ₹100 each month, this forms an Arithmetic Progression with first term $a = 500$ and common difference $d = 100$ .

Step 2: Find the savings in the 12th month

We use the $n^{\text{th}}$ term formula $a_n = a + (n - 1)d$ . Substituting $a = 500$ , $d = 100$ , and $n = 12$ , his savings in the 12th month is ₹1,600.

Step 3: Find the total amount saved in 12 months

We use the sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$ . Substituting $n = 12$ , $a = 500$ , and $d = 100$ , the total amount saved in 12 months is ₹12,600.

Step 4: Set up the inequality for total savings to exceed ₹20,000

For the total savings $S_n$ to exceed ₹20,000, we set $S_n > 20000$ . Simplifying the expression inside gives $\frac{n}{2}[1000 + 100n - 100] = \frac{n}{2}[100n + 900] = 50n(n + 9) > 20000$ , which reduces to $n(n + 9) > 400$ , or $n^2 + 9n - 400 > 0$ .

Step 5: Find the minimum number of months

Solving the equation $n^2 + 9n - 400 = 0$ by factoring, we get $(n + 25)(n - 16) = 0$ . Since $n$ must be a positive integer, $n = 16$ . At $n = 16$ , the total savings is exactly $16 \times 25 \times 50 = 20000$ rupees. Therefore, for the total savings to strictly exceed ₹20,000, the minimum number of months required is 17.

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