Algebra PYQ — Page 6

Let $\alpha=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\ldots\infty$ and $\beta=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ldots\infty$. Then the value of $(0.2)^{\log_{\sqrt{5}}(\alpha)}+(0.04)^{\log_5(\beta)}$ is equal to:

If $\displaystyle\sum_{k=1}^{n} a_k = 6n^3$, then $\displaystyle\sum_{k=1}^{6} \left(\dfrac{a_{k+1} - a_k}{36}\right)^2$ is equal to _______.

In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is $\mathbb{R}-(a, b)$, then $a^{2}+b^{2}$ is equal to $\_\_\_\_$.

Let the arithmetic mean of $\frac{1}{\mathrm{a}}$ and $\frac{1}{\mathrm{~b}}$ be $\frac{5}{16}, \mathrm{a}>2$. If $\alpha$ is such that $\mathrm{a}, 4, \alpha, \mathrm{~b}$ are in A.P., then the equation $\alpha x^{2}-a x+2(\alpha-2 b)=0$ has:

Let $a_{1}, a_{2}, a_{3}, a_{4}$ be an A.P. of four terms such that each term of the A.P. and its common difference $l$ are integers. If $a_{1}+a_{2}+a_{3}+a_{4}=48$ and $a_{1} a_{2} a_{3} a_{4}+l^{4}=361$, then the largest term of the A.P. is equal to

Let $a_{1}, \frac{a_{2}}{2}, \frac{a_{3}}{2^{2}}, \ldots, \frac{a_{10}}{2^{9}}$ be a G.P. of common ratio $\frac{1}{\sqrt{2}}$. If $a_{1}+a_{2}+\ldots+a_{10}=62$, then $a_{1}$ is equal to :

Let $\sum_{k=1}^{n} a_{k}=\alpha n^{2}+\beta n$. If $a_{10}=59$ and $a_{6}=7 a_{1}$, then $\alpha+\beta$ is equal to

If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is

Consider an A.P.: $a_{1}, a_{2}, \ldots, a_{\mathrm{n}} ; a_{1}>0$. If $a_{2}-a_{1}=\frac{-3}{4}, a_{\mathrm{n}}=\frac{1}{4} a_{1}$, and $\sum_{\mathrm{i}=1}^{\mathrm{n}} a_{\mathrm{i}}=\frac{525}{2}$, then $\sum_{\mathrm{i}=1}^{17} a_{\mathrm{i}}$ is equal to

Let $a_{1}=1$ and for $n \geqslant 1, a_{n+1}=\frac{1}{2} a_{n}+\frac{n^{2}-2 n-1}{n^{2}(n+1)^{2}}$. Then $\left|\sum_{n=1}^{\infty}\left(a_{n}-\frac{2}{n^{2}}\right)\right|$ is equal to $\_\_\_\_$.

Let $\alpha, \beta$ be the roots of the equation $x^2 - x + p = 0$ and $\gamma, \delta$ be the roots the equation $x^2 - 4x + q = 0$; $p, q \in \mathbf{Z}$. If $\alpha, \beta, \gamma, \delta$ are in G.P., then $|p + q|$ equals :

Given below are two statements: Statement I: The function $f: \mathbf{R} \rightarrow \mathbf{R}$ defined by $f(x)=\frac{x}{1+|x|}$ is one-one. Statement II: The function $f: \mathbf{R} \rightarrow \mathbf{R}$ defined by $f(x)=\frac{x^{2}+4 x-30}{x^{2}-8 x+18}$ is many-one. In the light of the above statements, choose the correct answer from the options given below :

The largest $\mathrm{n} \in \mathbf{N}$, for which $7^{\mathrm{n}}$ divides 101!, is :

The sum of all the roots of the equation $(x-1)^{2}-5|x-1|+6=0$, is :

A person has three different bags and four different books. The number of ways, in which he can put these books in the bags so that no bag is empty, is:

Let $a_{1}, a_{2}, a_{3}, \ldots$ be G.P. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4}=64$ and $a_{1}+a_{3}+a_{5}=\frac{813}{7}$. Then $a_{3}+a_{5}+a_{7}$ is equal to :

Let the circles $C_1 : |z| = r$ and $C_2 : |z - 3 - 4i| = 5$, $z \in \mathbb{C}$, be such that $C_2$ lies within $C_1$. If $z_1$ moves on $C_1$, $z_2$ moves on $C_2$ and $\min |z_1 - z_2| = 2$, then $\max |z_1 - z_2|$ is equal to:

Let $A = \begin{bmatrix} \alpha & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + \text{adj}(A)$. If $\det(B)=66$, then $\det(\text{adj}(A))$ equals:

The number of elements in the set $S = \left\{(r, k) : k \in \mathbb{Z} \text{ and } {}^{36}C_{r+1} = \dfrac{6\left({}^{35}C_r\right)}{(k^2 - 3)}\right\}$, is :

Let $f$ be a polynomial function such that $\log_2(f(x)) = \left(\log_2\left(2+\dfrac{2}{3}+\dfrac{2}{9}+\ldots\infty\right)\right)\cdot\log_3\left(1+\dfrac{f(x)}{f(1/x)}\right)$, $x>0$ and $f(6)=37$. Then $\displaystyle\sum_{n=1}^{10}f(n)$ is equal to ________.

The system of linear equations $x+y+z=6$ $2 x+5 y+a z=36$ $x+2 y+3 z=b$ has

Let $A$ be the set of first 101 terms of an A.P., whose first term is 1 and the common difference is 5 and let $B$ be the set of first 71 terms of an A.P., whose first term is 9 and the common difference is 7. Then the number of elements in $A \cap B$, which are divisible by 3, is :

The sum of the first ten terms of an A.P. is $160$ and the sum of the first two terms of a G.P. is $8$. If the first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to common difference of the A.P., then the sum of all possible values of the first term of the G.P. is:

Let $S=\left\{z \in \mathbb{C}: 4 z^{2}+\bar{z}=0\right\}$. Then $\sum_{z \in S}|z|^{2}$ is equal to: