The sum of the series $1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + \ldots$ upto $11 ^ { \text {th} }$ term is:
(1) 945
(2) 916
(3) 946
(4) 915

If $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots$ are in A.P. such that $a _ { 1 } + a _ { 7 } + a _ { 16 } = 40$, then the sum of the first 15 terms of this A.P. is

The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is
(1) 1356
(2) 1365
(3) 1256
(4) 1465

If the sum of the first 15 terms of the series $\left( \frac { 3 } { 4 } \right) ^ { 3 } + \left( 1 \frac { 1 } { 2 } \right) ^ { 3 } + \left( 2 \frac { 1 } { 4 } \right) ^ { 3 } + 3 ^ { 3 } + \left( 3 \frac { 3 } { 4 } \right) ^ { 3 } + \ldots$ is equal to 225 K , then $K$ is equal to :
(1) 9
(2) 27
(3) 54
(4) 108

Let the sum of the first $n$ terms of a non-constant A.P., $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots , a _ { n }$ be $50 n + \frac { n ( n - 7 ) } { 2 } A$, where $A$ is a constant. If $d$ is the common difference of this A.P., then the ordered pair $\left( d , a _ { 50 } \right)$ is equal to
(1) $( 50,50 + 46 A )$
(2) $( A , 50 + 45 A )$
(3) $( 50,50 + 45 A )$
(4) $( A , 50 + 46 A )$

Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 99 more identical balls are added to the total number of balls used in forming the equilateral triangle, then all these balls can be arranged in a square, whose each side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then the number of balls used to form the equilateral triangle is
(1) 262
(2) 190
(3) 225
(4) 157

Let $S _ { k } = \frac { 1 + 2 + 3 + \ldots + k } { k }$. If $S _ { 1 } ^ { 2 } + S _ { 2 } ^ { 2 } + \ldots + S _ { 10 } ^ { 2 } = \frac { 5 } { 12 } A$, then $A$ is equal to :
(1) 301
(2) 303
(3) 156
(4) 283

The sum $\sum _ { k = 1 } ^ { 20 } k \frac { 1 } { 2 ^ { k } }$ is equal to
(1) $1 - \frac { 11 } { 3 ^ { 20 } }$
(2) $2 - \frac { 21 } { 2 ^ { 20 } }$
(3) $2 - \frac { 3 } { 2 ^ { 17 } }$
(4) $2 - \frac { 11 } { 2 ^ { 19 } }$

If the sum and product of the first three terms in an A.P. are 33 and 1155, respectively, then a value of its $11^{\text{th}}$ term is:
(1) $- 25$
(2) $- 35$
(3) $25$
(4) $- 36$

If ${ } ^ { n } C _ { 4 } , { } ^ { n } C _ { 5 }$ and ${ } ^ { n } C _ { 6 }$ are in A.P., then $n$ can be
(1) 9
(2) 14
(3) 12
(4) 11

The sum of the series $2\cdot{}^{20}C_0 + 5\cdot{}^{20}C_1 + 8\cdot{}^{20}C_2 + 11\cdot{}^{20}C_3 + \ldots + 62\cdot{}^{20}C_{20}$ is equal to
(1) $2^{26}$
(2) $2^{25}$
(3) $2^{24}$
(4) $2^{23}$

If the lengths of the sides of a triangle are in A.P and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is:
(1) $3 : 4 : 5$
(2) $5 : 6 : 7$
(3) $5 : 9 : 13$
(4) $4 : 5 : 6$

If the $10^{\text{th}}$ term of an A.P. is $\frac{1}{20}$, and its $20^{\text{th}}$ term is $\frac{1}{10}$, then the sum of its first 200 terms is.
(1) 50
(2) $50\frac{1}{4}$
(3) 100
(4) $100\frac{1}{2}$

If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P. is
(1) $\frac { 1 } { 6 }$
(2) $\frac { 1 } { 5 }$
(3) $\frac { 1 } { 4 }$
(4) $\frac { 1 } { 7 }$

If the sum of the series $20 + 19 \frac { 3 } { 5 } + 19 \frac { 1 } { 5 } + 18 \frac { 4 } { 5 } + \ldots\ldots\ldots$ up to $n ^ { \text {th} }$ term is 488 and the $n ^ { \text {th} }$ term is negative, then :
(1) $n ^ { \text {th} }$ term is $- 4 \frac { 2 } { 5 }$
(2) $n = 41$
(3) $n ^ { \text {th} }$ term is - 4
(4) $n = 60$

Let $a _ { 1 } , a _ { 2 } , \ldots , a _ { n }$ be a given A.P. whose common difference is an integer and $S _ { n } = a _ { 1 } + a _ { 2 } + \ldots + a _ { n }$. If $a _ { 1 } = 1 , a _ { n } = 300$ and $15 \leq n \leq 50$, then the ordered pair $\left( \mathrm { S } _ { n - 4 } , a _ { n - 4 } \right)$ is equal to:
(1) $( 2490,249 )$
(2) $( 2480,249 )$
(3) $( 2480,248 )$
(4) $( 2490,248 )$

If $2 ^ { 10 } + 2 ^ { 9 } \cdot 3 ^ { 1 } + 2 ^ { 8 } \cdot 3 ^ { 2 } + \ldots\ldots + 2 \cdot 3 ^ { 9 } + 3 ^ { 10 } = S - 2 ^ { 11 }$, then $S$ is equal to
(1) $3 ^ { 11 } - 2 ^ { 12 }$
(2) $3 ^ { 11 }$
(3) $\frac { 3 ^ { 11 } } { 2 } + 2 ^ { 10 }$
(4) $2.3 ^ { 11 }$

The common difference of the A.P. $b_{1},b_{2},\ldots,b_{m}$ is 2 more than common difference of A.P. $\mathrm{a}_{1},\mathrm{a}_{2},\ldots,\mathrm{a}_{\mathrm{n}}$. If $\mathrm{a}_{40}=-159,\mathrm{a}_{100}=-399$ and $\mathrm{b}_{100}=\mathrm{a}_{70}$, then $\mathrm{b}_{1}$ is equal to:
(1) 81
(2) $-127$
(3) $-81$
(4) 127

Five numbers are in A.P., whose sum is 25 and product is 2520. If one of these five numbers is $- \frac { 1 } { 2 }$, then the greatest number amongst them is
(1) 27
(2) 7
(3) $\frac { 21 } { 2 }$
(4) 16

If the sum of the first 40 terms of the series, $3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + \ldots$. is $( 102 ) \mathrm { m }$, then m is equal to
(1) 20
(2) 25
(3) 5
(4) 10

If the sum of first 11 terms of an A.P. , $a _ { 1 } , a _ { 2 } , a _ { 3 } \ldots\ldots$ is $0 \left( a _ { 1 } \neq 0 \right)$ then the sum of the A.P $a _ { 1 } , a _ { 3 } , a _ { 5 } , \ldots\ldots a _ { 23 }$ is $k a _ { 1 }$ where $k$ is equal to
(1) $- \frac { 121 } { 10 }$
(2) $\frac { 121 } { 10 }$
(3) $\frac { 72 } { 5 }$
(4) $- \frac { 72 } { 5 }$

If $1 + \left( 1 - 2 ^ { 2 } \cdot 1 \right) + \left( 1 - 4 ^ { 2 } \cdot 3 \right) + \left( 1 - 6 ^ { 2 } \cdot 5 \right) + \ldots \ldots + \left( 1 - 20 ^ { 2 } \cdot 19 \right) = \alpha - 220 \beta$, then an ordered pair $( \alpha , \beta )$ is equal to:
(1) $( 10,97 )$
(2) $( 11,103 )$
(3) $( 10,103 )$
(4) $( 11,97 )$

If $3 ^ { 2 \sin 2 \alpha - 1 } , 14$ and $3 ^ { 4 - 2 \sin 2 \alpha }$ are the first three terms of an A.P. for some $\alpha$, then the sixth term of this A.P. is
(1) 66
(2) 81
(3) 65
(4) 78

Let $a , b , c , d$ and $p$ be non-zero distinct real numbers such that $\left( a ^ { 2 } + b ^ { 2 } + c ^ { 2 } \right) p ^ { 2 } - 2 ( a b + b c + c d ) p + \left( b ^ { 2 } + c ^ { 2 } + d ^ { 2 } \right) = 0$. Then
(1) $a , b , c$ are in A.P.
(2) $a , c , p$ are in G.P.
(3) $a , b , c , d$ are in G.P.
(4) $a , b , c , d$ are in A.P.

Let $S$ be the sum of the first 9 term of the series : $\{ x + k a \} + \left\{ x ^ { 2 } + ( k + 2 ) a \right\} + \left\{ x ^ { 3 } + ( k + 4 ) a \right\} + \left\{ x ^ { 4 } + ( k + 6 ) a \right\} + \ldots$ where $a \neq 0$ and $x \neq 1$. If $S = \frac { x ^ { 10 } - x + 45 a ( x - 1 ) } { x - 1 }$, then $k$ is equal to
(1) - 5
(2) 1
(3) - 3
(4) 3