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If $\alpha ,\beta \: \: and\: \: \gamma$ are three consecutive terms of a non-constant

G.P. such that the equations $\alpha x^{2}+2\beta x+\gamma =0$  and

$x^{2}+ x-1=0$ have a common root , then $\alpha (\beta +\gamma )$ is equal to :

• Option 1)

0

• Option 2)

$\alpha \beta$

• Option 3)

$\alpha \gamma$

• Option 4)

$\beta \gamma$

are in G.P. =>  For equation,                  Hence, roots are equal & equals to  Since, given equation have common roots , hence   must be root of                     Option 1) 0 Option 2) Option 3) Option 4)

If $a_1,a_2,a_3,.................$ 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 :

• Option 1)

200

• Option 2)

280

• Option 3)

120

• Option 4)

150

Given ,                 ..................(1) We have to find out , ..............(2) Substituting the value of (1) in (2), Option 1)200Option 2)280Option 3)120Option 4)150

if $\alpha \: and\: \beta$ are roots of the equation $375x^{2}-25x-2=0,$ then $\lim_{n\rightarrow \infty }\sum_{r=1}^{n}\alpha ^{r}+\lim_{n\rightarrow \infty }\sum_{r=1}^{n}\beta ^{r}$ is equal to :

• Option 1)

$\frac{1}{12}$

• Option 2)

$\frac{7}{116}$

• Option 3)

$\frac{21}{346}$

• Option 4)

$\frac{29}{358}$

Option 1) Option 2) Option 3) Option 4)

Let Sn denote the sum of the first terms of an A.P.. If $S_{4}=16$ and $S_{6}=-48$ then $S_{10}$ is equal to :

• Option 1)

$-380$

• Option 2)

$-320$

• Option 3)

$-260$

• Option 4)

$-410$

So (2) - (1)                                    Option 1) Option 2) Option 3) Option 4)

Let $a_1,a_2,a_3,.........$ be an A.P. with $a_6=2$ . Then the

common difference of this A.P., which maximises the product

$a_1a_4a_5$ , is :

• Option 1)

$\frac{3}{2}$

• Option 2)

• Option 3)

$\frac{6}{5}$

• Option 4)

$\frac{2}{3}$

Assuming the first term of A.P. is a and difference is d. Then, Let  =>  So,  will be maximum at  So, option (2) is correct. Option 1) Option 2) Option 3) Option 4)

The sum   $\mathrm{1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+..........+\frac{1^3+2^3+3^3+......+15^3}{1+2+3+.......+15}\:-\:\frac{1}{2}\left(1+2+3+.......+15\right)}$

is equal to :

• Option 1)

620

• Option 2)

1240

• Option 3)

1860

• Option 4)

660

So, option (1) is correct.   Option 1) 620 Option 2) 1240 Option 3) 1860 Option 4) 660

The sum

$\frac{3\times 1^{3}}{1^{2}}+\frac{5\times (1^{3}+2^{3})}{1^{2}+2^{2}}+\frac{7\times (1^{3}+2^{3}+3^{3})}{1^{2}+2^{2}+3^{2}}+..........$

upto 10th term , is :

• Option 1)

680

• Option 2)

600

• Option 3)

660

• Option 4)

620

Given, general term will be                                                               So, correct option is (3).Option 1)680Option 2)600Option 3)660Option 4)620

If $a_{1},a_{2},a_{3},..........a_{n}$ are in A.P. and $a_{1}+a_{4}+a_{7}..........+a_{16}=114$,

then $a_{1}+a_{6}+a_{11}+a_{16}$ is equal to :

• Option 1)

98

• Option 2)

76

• Option 3)

38

• Option 4)

64

correct option is (2). Option 1) 98 Option 2) 76 Option 3) 38 Option 4) 64

The sum of the series  $1+2\times3+3\times 5+4\times7+.......upto\:\:\:11^{th}$ term is :

• Option 1)

$915$

• Option 2)

$946$

• Option 3)

$945$

• Option 4)

$916$

Option 1) Option 2) Option 3) Option 4)

If the sum and  product  of the first three terms in an A.P. are $33$ and $1155$  , respectively , then a value of its $11^{th}$  term is :

• Option 1)

$-35$

• Option 2)

$25$

• Option 3)

$-36$

• Option 4)

$-25$

Let the terms be  or answer = 4Option 1)Option 2)Option 3)Option 4)

Some identical balls are arranged in rows to form an equilateral triangle . The first row consists of one ball , the second row consists of  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 :

• Option 1)

$157$

• Option 2)

$262$

• Option 3)

$225$

• Option 4)

$190$

let total number of rows in forming equilateral triangle is n  then  option 4 is answer   Option 1) Option 2) Option 3) Option 4)

Let $\inline \sum_{k=1}^{10}f\left ( a+k \right )=16\left ( 2^{10-1} \right )$  , where the function $\inline f$ satisfies $\inline f\left ( x+y \right )=f\left ( x \right )f\left ( y \right )$ for all natural numbers $\inline x,y$ and $\inline f\left ( 1 \right )=2 \; .$ Then the natural numbers $\inline 'a'$ is :

• Option 1)

$\inline 2$

• Option 2)

$\inline 16$

• Option 3)

$\inline 4$

• Option 4)

$\inline 3$

and   put                                                                                        Similarly   = Now,                                                                                                                                                           Option 1) Option 2) Option 3) Option 4)

If

$\begin{bmatrix} 1& 1\\ 0& 1 \end{bmatrix}$$\begin{bmatrix} 1& 2\\ 0& 1 \end{bmatrix}$$\begin{bmatrix} 1 & 3\\ 0 & 1 \end{bmatrix}.$.................$\begin{bmatrix} 1 & n-1\\ 0& 1 \end{bmatrix}=$$\begin{bmatrix} 1 & 78\\ 0& 1 \end{bmatrix}$ ,

then the inverse of $\begin{bmatrix} 1 & n\\ 0 & 1 \end{bmatrix}$ is :

• Option 1)

$\begin{bmatrix} 1 & 0\\ 12 & 1 \end{bmatrix}$

• Option 2)

$\begin{bmatrix} 1& -13\\ 0& 1\end{bmatrix}$

• Option 3)

$\begin{bmatrix} 1 & -12\\ 0 & 1 \end{bmatrix}$

• Option 4)

$\begin{bmatrix} 1 & 0\\ 13 &1 \end{bmatrix}$

and  and so on So,...........                  Inverse of   is   Option 1)    Option 2)   Option 3)   Option 4)

Let the sum of the first $n$ terms of a non-constant A.P.,$a_{1},a_{2},a_{3},.....$ be $50n+\frac{n\left ( n-7 \right )}{2}A,$ whrere $A$ 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 :

• Option 1)

$\left ( 50,50+46A \right )$

• Option 2)

$\left ( 50,50+45A \right )$

• Option 3)

$\left ( A,50+45A \right )$

• Option 4)

$\left ( A,50+46A \right )$

Given ,...................(1)   .........................(2)            Option 1) Option 2) Option 3)   Option 4)

Let the numbers $2,b,c$ be in an A.P. and $A=\begin{vmatrix} 1 &1 &1 \\ 2 &b &c \\ 4 &b^{2} &c^{2} \end{vmatrix}.$. If $\det (A)\epsilon \left [ 2,16 \right ],$ then c lies in the interval :

• Option 1)

$[2,3)$

• Option 2)

$\left ( 2+2^{\frac{3}{4}},4 \right )$

• Option 3)

$\left [ 4,6 \right ]$

• Option 4)

$\left [ 3,2+2^{\frac{3}{4}} \right ]$

are in A.P Option 1) Option 2) Option 3)   Option 4)

The sum $\sum_{k=1}^{20}k\frac{1}{2^{k}}$ is equal to :

• Option 1)

$2-\frac{3}{2^{17}}$

• Option 2)

$1-\frac{11}{2^{20}}$

• Option 3)

$2-\frac{11}{2^{19}}$

• Option 4)

$2-\frac{21}{2^{20}}$

Multiply  both side Substract (1) - (2) Option 1) Option 2) Option 3)   Option 4)

If three distinct numbers $a,b,c$  are in $G.P.$ and the equations $ax^{2}+2bx+c=0$ and $dx^{2}+2ex+f=0$ have a common root, then which one of the following statements is correct ?

• Option 1)

$\frac{d}{a},\frac{e}{b},\frac{f}{c}$  are in A.P.

• Option 2)

$d,e,f$ are in A.P.

• Option 3)

$d,e,f$ are in G.P.

• Option 4)

$\frac{d}{a},\frac{e}{b},\frac{f}{c}$ are in G.P.

Given  are in G.P. roots are =  root of this eq =  or     divide by ac Option 1)   are in A.P. Option 2)  are in A.P. Option 3)  are in G.P.   Option 4)  are in G.P.

The sum of all natural numbers 'n' such that $1001\; is :$

• Option 1)

$3121$

• Option 2)

$3203$

• Option 3)

$3303$

• Option 4)

$3221$

Natural number between 100 and 200 Number should either divide by 7 or divide by 13 (sum of no.divisible by 7)+(sum of no divisible by 13)-(sum of no divisible by 91) Option 1)Option 2)Option 3)  Option 4)

The sum of the squares of the lengths of the chords intercepted on the circle, $x^{2}+y^{2}=16$, by the lines $x+y=n,\; \; n\epsilon N,$ where $N$ is the set of all natural numbers, is :

• Option 1)

$105$

• Option 2)

$210$

• Option 3)

$160$

• Option 4)

$320$

then length of perpendicular from centre to line length of intercepts = Possible value of n are = Sum of squares of length = Option 1) Option 2) Option 3) Option 4)

Let $a_{1},a_{2},...........,a_{10}$  be a G.P. If  $\frac{a_{3}}{a_{1}}=25,$ then   $\frac{a_{9}}{a_{5}}$    equals :

• Option 1)

$4(5^{2})$

• Option 2)

$5^{4}$

• Option 3)

$2(5^{2})$

• Option 4)

$5^{3}$

Geometric Progession (GP) - A progression of non - zero terms, in which every term bears to the preceding term a constant ratio. - wherein eg 2, 4, 8, 16,- - - - - - and 100, 10, 1, 1/10,- - - - - - -     General term of a GP -   - wherein first term common ratio   Let first term of G.D be a and common ratio = r Now, Option 1)  Option 2)  Option 3)  Option 4)
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