If K = left|begin{array}{ll}3 & 4 5 & 4end{a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the $n$-th term of the series be $D_n = \left|\begin{array}{cc} a_n & b_n \ 5 & 4 \end{array}ight| = 4a_n - 5b_n$. Here, $a_n$ is a geometr

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let the $n$-th term of the series be $D_n = \left|\begin{array}{cc} a_n & b_n \ 5 & 4 \end{array}ight| = 4a_n - 5b_n$. Here, $a_n$ is a geometric progression with first term $a_1 = 3$ and common ratio $r_1 = 1/3$. Thus, $a_n = 3(1/3)^{n-1}$. Also, $b_n$ is a geometric progression with first term $b_1 = 4$ and common ratio $r_2 = -1/4$. Thus, $b_n = 4(-1/4)^{n-1}$. Therefore, $K = \sum_{n=1}^{\infty} (4a_n - 5b_n) = 4 \sum_{n=1}^{\infty} a_n - 5 \sum_{n=1}^{\infty} b_n$. Using the sum of an infinite geometric series formula $S = \frac{a}{1-r}$: $\sum_{n=1}^{\infty} a_n = \frac{3}{1 - 1/3} = \frac{3}{2/3} = \frac{9}{2}$. $\sum_{n=1}^{\infty} b_n = \frac{4}{1 - (-1/4)} = \frac{4}{5/4} = \frac{16}{5}$. Substituting these values into the expression for $K$: $K = 4 \left(\frac{9}{2}ight) - 5 \left(\frac{16}{5}ight) = 18 - 16 = 2$.

Similar Questions

If $A$ and $B$ are two square matrices of order $3$ such that $AB = A$ and $BA = B$,and matrices $X$ and $Y$ are defined as $X = A^4 + B^4$ and $Y = A^{10} + B^{10}$,then the matrix $X - Y$ is:

If $\omega (\neq 1)$ is a cube root of unity,then the value of the determinant $\left| \begin{array}{ccc} 1 & 1 + i + \omega^2 & \omega^2 \\ 1 - i & -1 & \omega^2 - 1 \\ -i & -i + \omega - 1 & -1 \end{array} \right|$ is equal to

Let $A$ and $B$ be $3 \times 3$ real matrices such that $A$ is a symmetric matrix and $B$ is a skew-symmetric matrix. Then the system of linear equations $(A^2 B^2 - B^2 A^2) X = 0$,where $X$ is a $3 \times 1$ column matrix of unknown variables and $0$ is a $3 \times 1$ null matrix,has:

Let $\alpha, \beta$ be the roots of the equation $ax^2+bx+c=0$, where $a, b, c$ are real. If $s_n = \alpha^n + \beta^n$ and $\left|\begin{array}{ccc}3 & 1+s_1 & 1+s_2 \\ 1+s_1 & 1+s_2 & 1+s_3 \\ 1+s_2 & 1+s_3 & 1+s_4\end{array}\right| = k \frac{(a+b+c)^2}{a^4}$, then $k =$

Let $A = \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix}$ and $B$ be a matrix such that $B(I - A) = I + A$. Then the sum of the diagonal elements of $B^T B$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module