The sum of the series $\frac { 1 } { 1 - 3 \cdot 1 ^ { 2 } + 1 ^ { 4 } } + \frac { 2 } { 1 - 3 \cdot 2 ^ { 2 } + 2 ^ { 4 } } + \frac { 3 } { 1 - 3 \cdot 3 ^ { 2 } + 3 ^ { 4 } } +\ldots$ up to 10 terms is
(1) $\frac { 45 } { 109 }$
(2) $- \frac { 45 } { 109 }$
(3) $\frac { 55 } { 109 }$
(4) $- \frac { 55 } { 109 }$

The $20 ^ { \text {th} }$ term from the end of the progression $20,19 \frac { 1 } { 4 } , 18 \frac { 1 } { 2 } , 17 \frac { 3 } { 4 } , \ldots , - 129 \frac { 1 } { 4 }$ is :-
(1) - 118
(2) - 110
(3) - 115
(4) - 100

Let three real numbers $a , b , c$ be in arithmetic progression and $a + 1 , b , c + 3$ be in geometric progression. If $a > 10$ and the arithmetic mean of $a , b$ and $c$ is 8 , then the cube of the geometric mean of $a , b$ and $c$ is
(1) 128
(2) 316
(3) 120
(4) 312

For $x \geqslant 0$, the least value of K , for which $4 ^ { 1 + x } + 4 ^ { 1 - x } , \frac { \mathrm {~K} } { 2 } , 16 ^ { x } + 16 ^ { - x }$ are three consecutive terms of an A.P., is equal to :
(1) 8
(2) 4
(3) 10
(4) 16

Let $2^{\text{nd}}$, $8^{\text{th}}$ and $44^{\text{th}}$ terms of a non-constant A.P. be respectively the $1^{\text{st}}$, $2^{\text{nd}}$ and $3^{\text{rd}}$ terms of G.P. If the first term of A.P. is 1 then the sum of first 20 terms is equal to-
(1) 980
(2) 960
(3) 990
(4) 970

A software company sets up $m$ number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of $m$ is equal to:
(1) 150
(2) 180
(3) 160
(4) 125

If $8 = 3 + \frac { 1 } { 4 } ( 3 + p ) + \frac { 1 } { 4 ^ { 2 } } ( 3 + 2 p ) + \frac { 1 } { 4 ^ { 3 } } ( 3 + 3 p ) + \ldots \infty$, then the value of $p$ is

If $\mathrm { S } ( x ) = ( 1 + x ) + 2 ( 1 + x ) ^ { 2 } + 3 ( 1 + x ) ^ { 3 } + \cdots + 60 ( 1 + x ) ^ { 60 } , x \neq 0$, and $( 60 ) ^ { 2 } \mathrm {~S} ( 60 ) = \mathrm { a } ( \mathrm { b } ) ^ { \mathrm { b } } + \mathrm { b }$, where $a , b \in N$, then $( a + b )$ equal to $\_\_\_\_$

An arithmetic progression is written in the following way

			2
	11	5	14	8	17
20		23		26		29

The sum of all the terms of the $10 ^ { \text {th} }$ row is $\_\_\_\_$

Let $S_n$ be the sum to $n$-terms of an arithmetic progression $3, 7, 11, \ldots$, if $40 < \frac{6}{n(n+1)}\sum_{k=1}^{n} S_k < 42$, then $n$ equals $\underline{\hspace{1cm}}$.

If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to
(1) - 1080
(2) - 1020
(3) - 1200
(4) - 120

Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its $11^{\text{th}}$ term is:
(1) 90
(2) 84
(3) 122
(4) 108

Let $S_n = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \ldots$ upto $n$ terms. If the sum of the first six terms of an A.P. with first term $-p$ and common difference $p$ is $\sqrt{2026 S_{2025}}$, then the absolute difference between $20^{\text{th}}$ and $15^{\text{th}}$ terms of the A.P. is
(1) 20
(2) 90
(3) 45
(4) 25

For positive integers $n$, if $4 a _ { n } = \left( n ^ { 2 } + 5 n + 6 \right)$ and $S _ { n } = \sum _ { k = 1 } ^ { n } \left( \frac { 1 } { a _ { k } } \right)$, then the value of $507 S _ { 2025 }$ is :
(1) 540
(2) 675
(3) 1350
(4) 135

Suppose that the number of terms in an A.P. is $2k , k \in N$. If the sum of all odd terms of the A.P. is 40 , the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27 , then k is equal to :
(1) 6
(2) 5
(3) 8
(4) 4

Let $\mathrm { T } _ { \mathrm { r } }$ be the $\mathrm { r } ^ { \text {th} }$ term of an A.P. If for some $\mathrm { m } , T _ { m } = \frac { 1 } { 25 } , T _ { 25 } = \frac { 1 } { 20 }$, and $20 \sum _ { \mathrm { r } = 1 } ^ { 25 } T _ { \mathrm { r } } = 13$, then $5 \mathrm { m } \sum _ { \mathrm { r } = \mathrm { m } } ^ { 2 \mathrm { m } } T _ { \mathrm { r } }$ is equal to
(1) 98
(2) 126
(3) 142
(4) 112

In an arithmetic progression, if $S_{40} = 1030$ and $S_{12} = 57$, then $S_{30} - S_{10}$ is equal to:
(1) 525
(2) 510
(3) 515
(4) 505

The roots of the quadratic equation $3 x ^ { 2 } - \mathrm { p } x + \mathrm { q } = 0$ are $10 ^ { \text {th} }$ and $11 ^ { \text {th} }$ terms of an arithmetic progression with common difference $\frac { 3 } { 2 }$. If the sum of the first 11 terms of this arithmetic progression is 88, then $q - 2 p$ is equal to

Let $a _ { 1 } , a _ { 2 } , \ldots , a _ { 2024 }$ be an Arithmetic Progression such that $a _ { 1 } + \left( a _ { 5 } + a _ { 10 } + a _ { 15 } + \ldots + a _ { 2020 } \right) + a _ { 2024 } = 2233$. Then $a _ { 1 } + a _ { 2 } + a _ { 3 } + \ldots + a _ { 2024 }$ is equal to $\_\_\_\_$

The interior angles of a polygon with $n$ sides, are in an A.P. with common difference $6 ^ { \circ }$. If the largest interior angle of the polygon is $219 ^ { \circ }$, then n is equal to

Q62. The value of $\frac { 1 \times 2 ^ { 2 } + 2 \times 3 ^ { 2 } + \ldots + 100 \times ( 101 ) ^ { 2 } } { 1 ^ { 2 } \times 2 + 2 ^ { 2 } \times 3 + \ldots + 100 ^ { 2 } \times 101 }$ is
(1) $\frac { 32 } { 31 }$
(2) $\frac { 31 } { 30 }$
(3) $\frac { 306 } { 305 }$
(4) $\frac { 305 } { 301 }$

Q62. If $\frac { 1 } { \sqrt { 1 } + \sqrt { 2 } } + \frac { 1 } { \sqrt { 2 } + \sqrt { 3 } } + \ldots + \frac { 1 } { \sqrt { 99 } + \sqrt { 100 } } = m$ and $\frac { 1 } { 1 \cdot 2 } + \frac { 1 } { 2 \cdot 3 } + \ldots + \frac { 1 } { 99 \cdot 100 } = n$, then the point $( \mathrm { m } , \mathrm { n } )$ lies on the line
(1) $11 ( x - 1 ) - 100 ( y - 2 ) = 0$
(2) $11 x - 100 y = 0$
(3) $11 ( x - 2 ) - 100 ( y - 1 ) = 0$
(4) $11 ( x - 1 ) - 100 y = 0$

Q62. If the sum of the series $\frac { 1 } { 1 \cdot ( 1 + \mathrm { d } ) } + \frac { 1 } { ( 1 + \mathrm { d } ) ( 1 + 2 \mathrm {~d} ) } + \ldots + \frac { 1 } { ( 1 + 9 \mathrm {~d} ) ( 1 + 10 \mathrm {~d} ) }$ is equal to 5 , then 50 d is equal to :
(1) 10
(2) 5
(3) 15
(4) 20

Q63. Let three real numbers $a , b , c$ be in arithmetic progression and $a + 1 , b , c + 3$ be in geometric progression. If $a > 10$ and the arithmetic mean of $a , b$ and $c$ is 8 , then the cube of the geometric mean of $a , b$ and $c$ is
(1) 128
(2) 316
(3) 120
(4) 312

Q63. For $x \geqslant 0$, the least value of K , for which $4 ^ { 1 + x } + 4 ^ { 1 - x } , \frac { \mathrm {~K} } { 2 } , 16 ^ { x } + 16 ^ { - x }$ are three consecutive terms of an A.P., is equal to :
(1) 8
(2) 4
(3) 10
(4) 16