JEE Main 2024 — 8 April, Shift 2 — Question 23

Step 1: Determine the starting term of the 10th row

Total number of terms before the $10^\text{th}$ row is:

$$\sum_{k=1}^{9} k = \frac{9 \cdot 10}{2} = 45$$

So, the first term of the 10th row is the $46^\text{th}$ term of the AP.

Using the formula for the $n^\text{th}$ term of an AP:

$$T_n = a + (n-1)d$$ $$T_{46} = 2 + (46 - 1) \cdot 3 = 2 + 135 = 137$$

Step 2: Determine the last term of the 10th row

Since the 10th row has 10 terms, the last term is:

$$T_{55} = 2 + (55 - 1) \cdot 3 = 2 + 162 = 164$$

Step 3: Use the sum formula for an AP

The sum of $n$ terms in an AP is:

$$S_n = \frac{n}{2}(a + l)$$

where $a$ is the first term, $l$ is the last term, and $n$ is the number of terms.

So, the sum of the 10th row is:

$$S = \frac{10}{2}(137 + 164) = 5 \cdot 301 = \boxed{1505}$$

Final Answer

$$\boxed{1505}$$