We use the n^(th) term formula of an AP, a_n = a + (n - 1)d. For the 10^(th) month, we substitute n = 10 and d = 50 to get a_10 = a + 9 × 50 = a + 450. If the student started with saving 50 (so a = 50), this would equal 500.

We use the n^(th) term formula of an AP, a_n = a + (n - 1)d. For the 10^(th) month, we substitute n = 10 and d = 50 to get a_10 = a + 9 × 50 = a + 450. If the student started with saving 50 (so a = 50), this would equal 500.

Using the same n^(th) term formula a_n = a + (n - 1)d for the 20^(th) month, we substitute n = 20 and d = 50. This yields a_20 = a + 19 × 50 = a + 950. If a = 50, the amount saved is 1000.

a student says a student saved in the first month and increases the amount saved by 50 every month find the amount saved in the 10th month amount saved in the 20th month calculate total amount saved in the first 20 months

Answer: In terms of the initial savings $a$ : - Amount saved in the $10^{\text{th}}$ month: $a + 450$ (or $500$ if $a = 50$ ) - Amount saved in the $20^{\text{th}}$ month: $a + 950$ (or $1000$ if $a = 50$ ) - Total amount saved in the first $20$ months: $20a + 9500$ (or $10500$ if $a = 50$ )

Step-by-step solution

Step 1: Analyze the given information and recognize the AP

Let the amount saved by the student in the first month be $a$ . Since the savings increase by a fixed amount of $50$ every month, the monthly savings form an Arithmetic Progression with common difference $d = 50$ . Notice that the problem statement does not provide a specific numerical value for the initial savings $a$ , which means the system is underdetermined for a single numerical answer; we express all required values in terms of the initial savings $a$ .

Step 2: Find the amount saved in the 10th month

We use the $n^{\text{th}}$ term formula of an AP, $a_n = a + (n - 1)d$ . For the $10^{\text{th}}$ month, we substitute $n = 10$ and $d = 50$ to get $a_{10} = a + 9 \times 50 = a + 450$ . If the student started with saving $50$ (so $a = 50$ ), this would equal $500$ .

Step 3: Find the amount saved in the 20th month

Using the same $n^{\text{th}}$ term formula $a_n = a + (n - 1)d$ for the $20^{\text{th}}$ month, we substitute $n = 20$ and $d = 50$ . This yields $a_{20} = a + 19 \times 50 = a + 950$ . If $a = 50$ , the amount saved is $1000$ .

Step 4: Calculate the total amount saved in the first 20 months

To calculate the total savings over the first $20$ months, we use the sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$ . Substituting $n = 20$ and $d = 50$ , we get $S_{20} = 10[2a + 950] = 20a + 9500$ . If the initial savings was $a = 50$ , the total sum would be $20(50) + 9500 = 10500$ .

Solve your own maths question free →