
Given the sequence:
\(2, 5, 11, 20, \ldots\)
Step 1: Finding the general term
The general term is given by:
\[
T_n = \frac{3n^2 - 3n + 4}{2}
\]
Step 2: Finding the 10th term
\[
T_{10} = \frac{3(100) - 3(10) + 4}{2}
\]
Simplifying:
\[
T_{10} = \frac{300 - 30 + 4}{2} = \frac{274}{2} = 137
\]
Hence, \(T_{10} = 137.\)
Step 3: Sum of 10 terms with common difference (c.d.) = 3
\[
S_{10} = \frac{10}{2} \left[ 2(137) + 9(3) \right]
\]
Simplifying:
\[
S_{10} = 5 (274 + 27) = 5 \times 301 = 1505
\]
Final Answer:
\[
S_{10} = 1505
\]
The sequence given is:
2, 5, 11, 20, …
The general term for the n-th row of this arithmetic progression can be expressed as:
\( T_n = \frac{3n^2 - 3n + 4}{2} \)
For the 10th row, we substitute \( n = 10 \):
\( T_{10} = \frac{3(100) - 3(10) + 4}{2} = \frac{300 - 30 + 4}{2} = \frac{274}{2} = 137 \)
Since there are 10 terms in the 10th row, with a common difference \( c.d. = 3 \), the sum of the terms of the 10th row is given by:
\( \text{Sum} = \frac{10}{2} (2 \times 137 + 9 \times 3) \)
Calculating:
\( \text{Sum} = 5 (274 + 27) = 5 \times 301 = 1505 \)
Answer: 1505
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of \( P_1 \) and \( P_2 \) are orthogonal to each other. The polarizer \( P_3 \) covers both the slits with its transmission axis at \( 45^\circ \) to those of \( P_1 \) and \( P_2 \). An unpolarized light of wavelength \( \lambda \) and intensity \( I_0 \) is incident on \( P_1 \) and \( P_2 \). The intensity at a point after \( P_3 \), where the path difference between the light waves from \( S_1 \) and \( S_2 \) is \( \frac{\lambda}{3} \), is:
