Multiple choice

If the sum of the first 𝑛 terms of an arithmetic progression is 4A whereas the sum of the first 2𝑛 terms of the same series is 11A, find the sum of the first 3𝑛 terms of the progression?

  1. 21A

  2. 15A

  3. 10A

  4. Cannot be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sum of n terms Sn = (n/2)(2a + (n-1)d) = 4A. Sum of 2n terms S2n = (2n/2)(2a + (2n-1)d) = 11A. 2a + (n-1)d = 8A/n. 2a + (2n-1)d = 11A/n. Subtracting: nd = 3A/n, so d = 3A/n^2. Then 2a = 8A/n - (n-1)(3A/n^2) = (8An - 3An + 3A)/n^2 = (5An + 3A)/n^2. S3n = (3n/2)(2a + (3n-1)d) = (3n/2)((5An+3A)/n^2 + (3n-1)(3A/n^2)) = (3n/2)(5An+3A + 9An-3A)/n^2 = (3n/2)(14An/n^2) = 21A.

AI explanation

The sum of the specific blocks of n terms follows an arithmetic progression because the overall sequence is an arithmetic progression. Let the first block sum be 4A, making the second block sum 11A minus 4A, which is 7A. The common difference between these block sums is 7A minus 4A, equaling 3A. Therefore, the third block sum equals 7A plus 3A, which is 10A, and adding this to the total of the first 2n terms (11A + 10A) gives the final result of 21A.

Keep practicing β€” related questions