
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
The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9 V with internal resistance of 1 \(\Omega\) is connected across these points is ______ J. 
The given circuit works as: 
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}