Let $a, b$ and $c$ be the $7^{\text{th}}, 11^{\text{th}}$ and $13^{\text{th}}$ terms respectively of a non-constant A.P. If these are also the three consecutive terms of a G.P., then $\frac{a}{c}$ is equal to:
(1) 2
(2) $\frac{7}{13}$
(3) $\frac{1}{2}$
(4) 4

$\lim _ { n \rightarrow \infty } \left( \frac { n } { n ^ { 2 } + 1 ^ { 2 } } + \frac { n } { n ^ { 2 } + 2 ^ { 2 } } + \frac { n } { n ^ { 2 } + 3 ^ { 2 } } + \ldots\ldots + \frac { 1 } { 5 n ^ { 2 } } \right)$ is equal to
(1) $\frac { \pi } { 4 }$
(2) $\tan ^ { - 1 } ( 2 )$
(3) $\frac { \pi } { 2 }$
(4) $\tan ^ { - 1 } ( 3 )$

Let $\alpha$ and $\beta$ be the roots of the equation $x ^ { 2 } - x - 1 = 0$. If $p _ { k } = ( \alpha ) ^ { k } + ( \beta ) ^ { k } , k \geq 1$, then which one of the following statements is not true?
(1) $p _ { 3 } = p _ { 5 } - p _ { 4 }$
(2) $p _ { 5 } = 11$
(3) $\left( p _ { 1 } + p _ { 2 } + p _ { 3 } + p _ { 4 } + p _ { 5 } \right) = 26$
(4) $p _ { 5 } = p _ { 2 } \cdot p _ { 3 }$

Let $\alpha$ and $\beta$ be the roots of the equation, $5x^{2} + 6x - 2 = 0$. If $S_{n} = \alpha^{n} + \beta^{n}, n = 1,2,3,\ldots$, then
(1) $6S_{6} + 5S_{5} = 2S_{4}$
(2) $5S_{6} + 6S_{5} + 2S_{4} = 0$
(3) $5S_{6} + 6S_{5} = 2S_{4}$
(4) $6S_{6} + 5S_{5} + 2S_{4} = 0$

The value of $4 + \frac { 1 } { 5 + \frac { 1 } { 4 + \frac { 1 } { 5 + \frac { 1 } { 4 + \ldots . . \infty } } } }$ is:
(1) $2 + \frac { 2 } { 5 } \sqrt { 30 }$
(2) $2 + \frac { 4 } { \sqrt { 5 } } \sqrt { 30 }$
(3) $4 + \frac { 4 } { \sqrt { 5 } } \sqrt { 30 }$
(4) $5 + \frac { 2 } { 5 } \sqrt { 30 }$

The value of $3 + \frac { 1 } { 4 + \frac { 1 } { 3 + \frac { 1 } { 4 + \frac { 1 } { 3 + \ldots . \infty } } } }$ is equal to
(1) $1.5 + \sqrt { 3 }$
(2) $2 + \sqrt { 3 }$
(3) $3 + 2 \sqrt { 3 }$
(4) $4 + \sqrt { 3 }$

Let $\alpha$ and $\beta$ be the roots of $x ^ { 2 } - 6 x - 2 = 0$. If $a _ { n } = \alpha ^ { n } - \beta ^ { n }$ for $n \geqslant 1$, then the value of $\frac { a _ { 10 } - 2 a _ { 8 } } { 3 a _ { 9 } }$ is:
(1) 1
(2) 3
(3) 2
(4) 4

The sum of the series $\sum _ { n = 1 } ^ { \infty } \frac { n ^ { 2 } + 6 n + 10 } { ( 2 n + 1 ) ! }$ is equal to
(1) $\frac { 41 } { 8 } e + \frac { 19 } { 8 } e ^ { - 1 } + 10$
(2) $\frac { 41 } { 8 } e + \frac { 19 } { 8 } e ^ { - 1 } - 10$
(3) $- \frac { 41 } { 8 } e + \frac { 19 } { 8 } e ^ { - 1 } - 10$
(4) $\frac { 41 } { 8 } e - \frac { 19 } { 8 } e ^ { - 1 } - 10$

The sum of the series $\frac { 1 } { x + 1 } + \frac { 2 } { x ^ { 2 } + 1 } + \frac { 2 ^ { 2 } } { x ^ { 4 } + 1 } + \ldots + \frac { 2 ^ { 100 } } { x ^ { 2 ^ { 100 } } + 1 }$ when $x = 2$ is:
(1) $1 - \frac { 2 ^ { 101 } } { 4 ^ { 101 } - 1 }$
(2) $1 + \frac { 2 ^ { 101 } } { 4 ^ { 101 } - 1 }$
(3) $1 + \frac { 2 ^ { 100 } } { 4 ^ { 101 } - 1 }$
(4) $1 - \frac { 2 ^ { 100 } } { 4 ^ { 201 } - 1 }$

If $0 < x < 1$ and $y = \frac { 1 } { 2 } x ^ { 2 } + \frac { 2 } { 3 } x ^ { 3 } + \frac { 3 } { 4 } x ^ { 4 } + \ldots \ldots$, then the value of $e ^ { 1 + y }$ at $x = \frac { 1 } { 2 }$ is: (1) $\frac { 1 } { 2 } e ^ { 2 }$ (2) $2 e$ (3) $2 e ^ { 2 }$ (4) $\frac { 1 } { 2 } \sqrt { \mathrm { e } }$

If $0 < a , b < 1$, and $\tan ^ { - 1 } a + \tan ^ { - 1 } b = \frac { \pi } { 4 }$, then the value of $( a + b ) - \left( \frac { a ^ { 2 } + b ^ { 2 } } { 2 } \right) + \left( \frac { a ^ { 3 } + b ^ { 3 } } { 3 } \right) - \left( \frac { a ^ { 4 } + b ^ { 4 } } { 4 } \right) + \ldots$ is :
(1) $\log _ { \mathrm { e } } \left( \frac { e } { 2 } \right)$
(2) $e$
(3) $e ^ { 2 } - 1$
(4) $\log _ { e } 2$

$\lim _ { n \rightarrow \infty } \left( 1 + \frac { 1 + \frac { 1 } { 2 } + \ldots\ldots + \frac { 1 } { n } } { n ^ { 2 } } \right) ^ { n }$ is equal to
(1) $\frac { 1 } { e }$
(2) 0
(3) $\frac { 1 } { 2 }$
(4) 1

Let $S _ { k } = \sum _ { r = 1 } ^ { k } \tan ^ { - 1 } \left( \frac { 6 ^ { r } } { 2 ^ { 2 r + 1 } + 3 ^ { 2 r + 1 } } \right)$, then $\lim _ { k \rightarrow \infty } S _ { k }$ is equal to :
(1) $\tan ^ { - 1 } \left( \frac { 3 } { 2 } \right)$
(2) $\frac { \pi } { 2 }$
(3) $\cot ^ { - 1 } \left( \frac { 3 } { 2 } \right)$
(4) $\tan ^ { - 1 } ( 3 )$

Consider an arithmetic series and a geometric series having four initial terms from the set $\{ 11,8,21,16,26,32,4 \}$. If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these two series is equal to $\_\_\_\_$.

Let $\left\{ a _ { n } \right\} _ { n = 1 } ^ { \infty }$ be a sequence such that $a _ { 1 } = 1 , a _ { 2 } = 1$ and $a _ { n + 2 } = 2 a _ { n + 1 } + a _ { n }$ for all $n \geq 1$. Then the value of $47 \sum _ { n = 1 } ^ { \infty } \left( \frac { a _ { n } } { 2 ^ { 3 n } } \right)$ is equal to $\underline{\hspace{1cm}}$.

If $A = \sum _ { n = 1 } ^ { \infty } \frac { 1 } { \left( 3 + ( - 1 ) ^ { n } \right) ^ { n } }$ and $B = \sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { \left( 3 + ( - 1 ) ^ { n } \right) ^ { n } }$, then $\frac { A } { B }$ is equal to
(1) $\frac { 11 } { 9 }$
(2) 1
(3) $- \frac { 11 } { 9 }$
(4) $- \frac { 11 } { 3 }$

Consider the sequence $a_1, a_2, a_3, \ldots$ such that $a_1 = 1$, $a_2 = 2$ and $a_{n+2} = \frac{2}{a_{n+1}} + a_n$ for $n = 1, 2, 3, \ldots$ If $\frac{a_1 + \frac{1}{a_2}}{a_3} \cdot \frac{a_2 + \frac{1}{a_3}}{a_4} \cdot \frac{a_3 + \frac{1}{a_4}}{a_5} \cdots \frac{a_{30} + \frac{1}{a_{31}}}{a_{32}} = 2^\alpha \binom{61}{31}$ then $\alpha$ is equal to
(1) $-30$
(2) $-31$
(3) $-60$
(4) $-61$

Let $\{a_n\}_{n=0}^{\infty}$ be a sequence such that $a_0 = a_1 = 0$ and $a_{n+2} = 3a_{n+1} - 2a_n + 1, \forall n \geq 0$. Then $a_{25}a_{23} - 2a_{25}a_{22} - 2a_{23}a_{24} + 4a_{22}a_{24}$ is equal to
(1) 483
(2) 528
(3) 575
(4) 624

Let $S = 2 + \frac { 6 } { 7 } + \frac { 12 } { 7 ^ { 2 } } + \frac { 20 } { 7 ^ { 3 } } + \frac { 30 } { 7 ^ { 4 } } + \ldots$. then $4 S$ is equal to
(1) $\left( \frac { 7 } { 2 } \right) ^ { 2 }$
(2) $\left( \frac { 7 } { 3 } \right) ^ { 3 }$
(3) $\frac { 7 } { 3 }$
(4) $\left( \frac { 7 } { 3 } \right) ^ { 4 }$

If $\lim _ { n \rightarrow \infty } \left( \sqrt { n ^ { 2 } - n - 1 } + n\alpha + \beta \right) = 0$ then $8\alpha + \beta$ is equal to
(1) 4
(2) - 8
(3) - 4
(4) 8

$\lim _ { n \rightarrow \infty } \left( \frac { n ^ { 2 } } { \left( n ^ { 2 } + 1 \right) ( n + 1 ) } + \frac { n ^ { 2 } } { \left( n ^ { 2 } + 4 \right) ( n + 2 ) } + \frac { n ^ { 2 } } { \left( n ^ { 2 } + 9 \right) ( n + 3 ) } + \ldots + \frac { n ^ { 2 } } { \left( n ^ { 2 } + n ^ { 2 } \right) ( n + n ) } \right)$ is equal to
(1) $\frac { \pi } { 8 } + \frac { 1 } { 4 } \ln 2$
(2) $\frac { \pi } { 4 } + \frac { 1 } { 8 } \ln 2$
(3) $\frac { \pi } { 4 } - \frac { 1 } { 8 } \ln 2$
(4) $\frac { \pi } { 8 } + \ln \sqrt { 2 }$

If $\lim _ { n \rightarrow \infty } \frac { (n+1)^{k-1} } { n ^ { k + 1 } } \left[ (nk+1) + (nk+2) + \ldots + (nk+n) \right] = 33 \cdot \lim _ { n \rightarrow \infty } \frac { 1 } { n ^ { k + 1 } } \cdot \left( 1 ^ { k } + 2 ^ { k } + 3 ^ { k } + \ldots + n ^ { k } \right)$, then the integral value of $k$ is equal to $\_\_\_\_$.

If $S _ { n } = 4 + 11 + 21 + 34 + 50 + \ldots$ to $n$ terms, then $\frac { 1 } { 60 } \left( S _ { 29 } - S _ { 9 } \right)$ is equal to
(1) 223
(2) 226
(3) 220
(4) 227

Let $a _ { n }$ be $n ^ { \text {th} }$ term of the series $5 + 8 + 14 + 23 + 35 + 50 + \ldots\ldots$. and $S _ { n } = \sum _ { k = 1 } ^ { n } a _ { k }$. Then $S _ { 30 } - a _ { 40 }$ is equal to
(1) 11310
(2) 11260
(3) 11290
(4) 11280

Let $\left\langle a _ { n } \right\rangle$ be a sequence such that $a _ { 1 } + a _ { 2 } + \ldots + a _ { n } = \frac { n ^ { 2 } + 3 n } { ( n + 1 ) ( n + 2 ) }$. If $28 \sum _ { k = 1 } ^ { 10 } \frac { 1 } { a _ { k } } = p _ { 1 } p _ { 2 } p _ { 3 } \ldots p _ { m }$, where $p _ { 1 } , p _ { 2 } , \ldots p _ { m }$ are the first $m$ prime numbers, then $m$ is equal to
(1) 5
(2) 8
(3) 6
(4) 7