What is the sum of the first 15 terms of an arithmetic progression where the fir

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Question: What is the sum of the first 15 terms of an arithmetic progression where the first term is 10 and the common difference is 2?

Options:

Correct Answer: 160

Solution:

The sum of the first n terms S_n = n/2 * (2a + (n-1)d). Here, S_15 = 15/2 * (2*10 + 14*2) = 15/2 * (20 + 28) = 15/2 * 48 = 360.

What is the sum of the first 15 terms of an arithmetic progression where the first term is 10 and the common difference is 2?

What is the sum of the first 15 terms of an arithmetic progression where the first term is 10 and the common difference is 2?

Step 1: Identify the first term (a) of the arithmetic progression. Here, a = 10.
Step 2: Identify the common difference (d) of the arithmetic progression. Here, d = 2.
Step 3: Identify the number of terms (n) you want to sum. Here, n = 15.
Step 4: Use the formula for the sum of the first n terms of an arithmetic progression: S_n = n/2 * (2a + (n-1)d).
Step 5: Substitute the values into the formula: S_15 = 15/2 * (2*10 + (15-1)*2).
Step 6: Calculate (15-1) which is 14, so now we have S_15 = 15/2 * (2*10 + 14*2).
Step 7: Calculate 2*10 which is 20 and 14*2 which is 28, so now we have S_15 = 15/2 * (20 + 28).
Step 8: Add 20 and 28 to get 48, so now we have S_15 = 15/2 * 48.
Step 9: Multiply 15 by 48 to get 720, then divide by 2 to get 360.
Step 10: The sum of the first 15 terms is 360.

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