We use the sum formula S_n = (n)/(2)[2a + (n - 1)d]. For n = 12, substituting a = 500 and d = 100 gives S_12 = 6 × 2100 = 12600. Thus, the total amount saved in 12 months is ₹ 12600.

Let n be the number of months required for total savings S_n to exceed ₹ 20000. Expanding the sum gives (n)/(2)[1000 + 100n - 100] > 20000, which simplifies to 50n(n + 9) > 20000, or n² + 9n - 400 > 0.

Let n be the number of months required for total savings S_n to exceed ₹ 20000. Expanding the sum gives (n)/(2)[1000 + 100n - 100] > 20000, which simplifies to 50n(n + 9) > 20000, or n² + 9n - 400 > 0.

a man say 500 rupees in the first month and increases his monthly savings by 100 every month fine he is savings in 12 month the total amounts saved in 12 months the minimum number of months required for his daughter saving to exceed 20000 rupees

Answer: (i) Savings in the 12th month = ₹ 1600, (ii) Total savings in 12 months = ₹ 12600, (iii) Minimum number of months to exceed ₹ 20000 = 17 months.

Step-by-step solution

Step 1: Identify the Arithmetic Progression parameters

The savings in consecutive months are $500, 600, 700, \dots$ . Since the difference between consecutive monthly savings is a constant ₹ $100$ , the savings form an arithmetic progression with first term $a = 500$ and common difference $d = 100$ .

Step 2: Find his savings in the 12th month

We use the $n^{\text{th}}$ term formula $a_n = a + (n - 1)d$ . For $n = 12$ , substituting $a = 500$ and $d = 100$ gives $a_{12} = 500 + 1100 = 1600$ . Therefore, his savings in the 12th month is ₹ $1600$ .

Step 3: Calculate the total amount saved in 12 months

We use the sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$ . For $n = 12$ , substituting $a = 500$ and $d = 100$ gives $S_{12} = 6 \times 2100 = 12600$ . Thus, the total amount saved in 12 months is ₹ $12600$ .

Step 4: Set up the inequality for total savings to exceed ₹ 20000

Let $n$ be the number of months required for total savings $S_n$ to exceed ₹ $20000$ . Expanding the sum gives $\frac{n}{2}[1000 + 100n - 100] > 20000$ , which simplifies to $50n(n + 9) > 20000$ , or $n^2 + 9n - 400 > 0$ .

Step 5: Determine the minimum integer value of n

Solving the equation $n^2 + 9n - 400 = 0$ using the quadratic formula gives $n = \frac{-9 + 41}{2} = 16$ (since $n > 0$ ). At $n = 16$ , the total savings is exactly $S_{16} = 8[1000 + 1500] = 20000$ . Since the savings must strictly exceed ₹ $20000$ , we require $n > 16$ , which means the minimum integer number of months is $17$ .

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