An arithmetic progression is written in the following way
2; 5, 8; 11, 14, 17; 20, 23, 26, 29; - - -, - - -, - - -, - - -, - - -, - - -, - - -; - - -, - - -, - - -, - - -, - - -, - - -, - - -
The sum of all the terms of the 10ᵗʰ row is _________.

Solution

Maths sir · 4 steps

Row n holds n terms, so first count how many terms come before row 10, then add that stretch of the AP.

1. Spot the AP. Read left to right, row by row: 2, 5, 8, 11, … with first term 2 and common difference 3. So the kth term is aₖ = 2 + 3(k-1) = 3k - 1.
2. Locate row 10. Row 1 has 1 term, row 2 has 2 terms, and so on. Rows 1 to 9 use up 1 + 2 + … + 9 = 45 terms, so row 10 holds the 10 terms from a₄₆ to a₅₅.
3. End terms. a₄₆ = 3 × 46 - 1 = 137 and a₅₅ = 3 × 55 - 1 = 164.

   S = 10/2(137 + 164) = 5 × 301 = 1505

4. So. The sum of all the terms in the 10th row is 1505.

The common slip: 1475. Starting row 10 at the 45th term instead of the 46th gives 1475. The nine rows before it use up all of terms 1 to 45.

The idea to keep. In a triangular arrangement, row n ends at term n(n+1)/2. Find where the row starts and ends, then use (number of terms)/2 × (first + last).

Answer from the NTA final answer key. Solution by PrepWiser.