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If 19th term of a non-zero A.P. is zero, then its ( 49th term ) : ( 29th term ) is :

• Option 1)

2:1

• Option 2)

4:1

• Option 3)

3:1

• Option 4)

1:3

General term of an A.P. - - wherein First term number of term common difference Option 1)2:1Option 2)4:1Option 3)3:1Option 4)1:3

Let x,y be positive real numbers and m,n positive integers. The maximum value of the expression $\frac{x^{m}y^{n}}{(1+x^{2m})(1+y^{2n})}$  is :

• Option 1)

1

• Option 2)

$\frac{1}{2}$

• Option 3)

$\frac{1}{4}$

• Option 4)

$\frac{m+n}{6mn}$

Relation between AM, GM and HM of two positive numbers - - wherein Inequality of the three given means.    1 Option 1)  1Option 2)Option 3)  Option 4)

The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is $\frac{27}{19}.$

Then the common ratio of this series is :

• Option 1)

$\frac{4}{9}$

• Option 2)

$\frac{1}{3}$

• Option 3)

$\frac{2}{9}$

• Option 4)

$\frac{2}{3}$

Sum of infinite terms of a GP - - wherein first term common ratio   Lrt first term of G.P is a and the common ratio is r (r<1) Given,    2)  Given,   2)  From (1) and (2) since r<1  Option 1)  Option 2)  Option 3)  Option 4)

If the sum of the first 15 terms of the series $\left ( \frac{3}{4} \right )^{3}+\left ( 1\frac{1}{2} \right )^{3}+\left ( 2\frac{1}{4} \right )^{3}+3^{3}+\left ( 3\frac{3}{4} \right )^{3}+\cdots$

is equal to 225 k, then k is equal to :

• Option 1)

$9$

• Option 2)

$27$

• Option 3)

$108$

• Option 4)

$54$

Summation of series of natural numbers -   - wherein Sum of first n natural numbers     Summation of series of natural numbers - - wherein        Option 1)  Option 2)  Option 3)  Option 4)

If $_{}^{n}\textrm{C}_{4}$ , $_{}^{n}\textrm{C}_{5}$ , and $_{}^{n}\textrm{C}_{6}$  are in A.P., then n can be :

• Option 1)

$12$

• Option 2)

$9$

• Option 3)

$14$

• Option 4)

$11$

Arithmetic mean of two numbers (AM) - - wherein It is to be noted that the sequence a, A, b, is in AP where, a and b are the two numbers.  Option 1)Option 2)Option 3)Option 4)

Let $a_1, a_2, a_3, ..., a_{10}$ be in GP with $a_i > 0$ for $i = 1,2,..., 10$ and S be the set of pairs (r, k), r, k$\in N$( the set of natural numbers) for which

$\begin{vmatrix} \log_e a_1^ra_2^k & \log_e a_2^ra_3^k &\log_e a_3^ra_4^k \\ \log_e a_4^ra_5^k & \log_e a_5^ra_6^k & \log_e a_6^ra_7^k \\ \log_e a_7^ra_8^k &\log_e a_8^ra_9^k &\log_e a_9^ra_{10}^k \end{vmatrix} = 0$

Then the number of elements in S, is:

• Option 1)

4

• Option 2)

Infinitely many

• Option 3)

10

• Option 4)

2

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 Apply coloumn operation  we get D = 0 OR    are in G.P. assume  Since  are in G.P. with common ratio 1  So,  Value of D become...

The sum of all two digit positive numbers which when divided by 7 yeild 2 or5 as remainder is  :

• Option 1)

1256

• Option 2)

1465

• Option 3)

1365

• Option 4)

1356

Summation of series of natural numbers -   - wherein Sum of first n natural numbers From the concept , General term will be  total = 1356Option 1)1256Option 2)1465Option 3)1365Option 4)1356

Let $S_{k}=\frac{1+2+3+\cdots +k}{k}.$  If ${S_{1}}^{2}+{S_{2}}^{2}+\cdots +{S_{10}}^{2}=\frac{5}{12}A,$  then A is equal to :

• Option 1)

$283$

• Option 2)

$301$

• Option 3)

$156$

• Option 4)

$303$

Summation of series of natural numbers -   - wherein Sum of first n natural numbers Since     Option 1)Option 2)Option 3)Option 4)

The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an A.P. Then the sum of the original three terms of the given G.P. is :

• Option 1)

$24$

• Option 2)

$32$

• Option 3)

$36$

• Option 4)

$28$

Selection of terms in G.P. - If we have to take three terms in GP, we take them as - wherein Extension : If we have to take (2K+1) term in GP, we take them as     General term of an A.P. - - wherein First term number of term common difference    be three terms. Putting  Numbers are 4,8,16 or 16, 8,4  Sum of numbers = 4+ 8 + 16 = 28Option 1)Option 2)Option 3)Option 4)

Let a,b and c be the 7th,11th and 13th terms respectively of a non-constant A.P.If these are also the three consecutive terms of a G.P., then a/c is equal to :

• Option 1)

2

• Option 2)

7/13

• Option 3)

1/2

• Option 4)

4

General term of an A.P. - - wherein First term number of term common difference       Selection of terms in G.P. - If we have to take three terms in GP, we take them as - wherein Extension : If we have to take (2K+1) term in GP, we take them as From the concept  and  and  are in G.P    Option 1)  2Option 2)  7/13Option 3)  1/2Option 4)  4

Let S be the set of all triangles in the xy-plane,each having one vertex at the origin and the other two vertices  lie on coordinate axes with integral coordinates. If each triangles is S has area 50 sq. units, then the number of the elements in the set S is:

• Option 1)

9

• Option 2)

32

• Option 3)

18

• Option 4)

36

Number of Divisions - The number of divisors of a natural number   is  - wherein Where a1, a2 ....... are distinct prime and non negative integers.     Let  and  be vectors of  Area of  le is  Number of triangles   Option 1)  9Option 2)  32Option 3)  18Option 4)  36

The sum of the following series

$1+6+\frac{9(1^{2}+2^{2}+3^{2})}{7}+\frac{12(1^{2}+2^{2}+3^{2}+4^{2})}{9}+\frac{15(1^{2}+2^{2}+3^{2}+.....+5^{2}}{11})+.......$

up to 15 terms, is:

• Option 1)

7830

• Option 2)

7820

• Option 3)

7510

• Option 4)

7520

Summation of series of natural numbers -   - wherein Sum of first n natural numbers     Summation of series of natural numbers -   - wherein Sum of  squares of first n natural numbers     Summation of series of natural numbers - - wherein   The general term of the sequence will be Option 1)  7830Option 2)  7820Option 3)  7510Option 4)  7520

Let $a_1, a_2,... ....., a_{30}$ be an AP, $S = \sum_{i = 1}^{30}a_i$ and  $T = \sum_{i = 1}^{15}a_{(2i-1)}$ .

If $a_5 = 27$ and $S-2T = 75$

then $a_{10}$ is equal to:

• Option 1)

52

• Option 2)

57

• Option 3)

47

• Option 4)

42

Sum of n terms of an AP -   and Sum of n terms of an AP   - wherein first term common difference number of terms   Last term of an A.P.(l) - - wherein If a series has n terms last term is   Summation of terms of an AP       or,  Now,            Now use, Give  Now,                  Option 1) 52Option 2)57Option 3)47Option 4)42

If a, b and c be three distinct real numbers in G.P and $a+b+c = xb$ then $x$ cannot be:

• Option 1)

-2

• Option 2)

-3

• Option 3)

4

• Option 4)

2

General term of a GP -   - wherein first term common ratio   from the concept we have learnt  Let the three terms be  given    from the AM-GM         So,   So, Option 1)-2Option 2)-3Option 3)4Option 4)2
Engineering
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If the sum of the first 15 terms of the series 3+7+14+24+37+.......is 15K, then k is equal to :

• Option 1)

126

• Option 2)

122

• Option 3)

81

• Option 4)

119

Use Summation of series of natural numbers -   - wherein Sum of first n natural numbers    and   Summation of series of natural numbers -   - wherein Sum of  squares of first n natural numbers     3+7+14+24+37 ----------15K Let Tn=an2+bn+c            because difference is in A.P: [4,7,10,13-----------------] at n=1     ------(i)     n=2      -------(ii)     n=3        -------(iii)    ...
Engineering
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Let G be the geometric mean of two positive numbers a and b, and M be the arithmetic mean of    and

if   is 4:5 then a:b can be:

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we learnt in

Arithmetic mean of two numbers (AM) -

$A=\frac{a+b}{2}$

- wherein

It is to be noted that the sequence a, A, b, is in AP where, a and b are the two numbers.

and

Geometric mean of two numbers (GM) -

$GM= \sqrt{ab}$

- wherein

It is to be noted that a,G,b are in GP and a,b are two non - zero numbers.

Given G= ab

$2M=\frac{1}{a}+\frac{1}{b}$

and  $\frac{\frac{1}{M}}{G}=\frac{4}{5}$

$\therefore GM=\frac{5}{4}$

$\therefore G^{2}M^{2}=\frac{25}{16}$

$=ab\times \left (\frac{a+b}{2ab} \right )^{2}= \frac{25}{16}$

$=\frac{ab\times \left ( a+b \right )^{2}}{4a^{2}b^{2}}= \frac{25}{16}$

$\therefore 4a^{2}+4b^{2}-17ab=0$

$\therefore 4\left ( \frac{a}{b} \right )^{2}-17\left ( \frac{a}{b} \right )+4=0$

$\therefore \frac{a}{b}=1:4$

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
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The sum of the first 20 terms common between the series 3+7+11+15+..... and 1+6+11+16+..... , is :

• Option 1)

4000

• Option 2)

4020

• Option 3)

4200

• Option 4)

4220

Use

Sum of n terms of an AP -

$S_{n}= \frac{n}{2}\left [ 2a +\left ( n-1 \right )d\right ]$

and

Sum of n terms of an AP

$S_{n}= \frac{n}{2}\left [ a+l\right ]$

- wherein

$a\rightarrow$ first term

$d\rightarrow$ common difference

$n\rightarrow$ number of terms

series are 3,7,11,15,19,23,27,31..............

and 1,6,11,16,21,26,31....................

So common terms are 11,31,............

$\therefore S_{20}= \frac{20}{2}\left [ 2\times 11+\left ( 20-1 \right )\times 20 \right ]$

$=10\left [ 22+380 \right ]$

$=10\left [402]$

$=4020$

Option 1)

4000

Option 2)

4020

Option 3)

4200

Option 4)

4220

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Engineering
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If the arithmetic mean of two numbers a and b, a > b > 0, is five times their geometric

mean, then $\frac{a+b}{a-b}$     is equal to:

• Option 1)

• Option 2)

• Option 3)

• Option 4)

Valve surrounding opening of the coronary sinus is.........

A) eustachian valve    b)  thebasious valve

C)mitral valve               d)tricuspid  valve

Engineering
56 Views   |

If (10)9 + 2(11)1   (10)8 + 3(11)2  (10)7 +......  +10 (11)9 = k (10)9, then k is equal to :

• Option 1)

100

• Option 2)

110

• Option 3)

• Option 4)

Use

Sum of n terms of a GP -

$S_{n}= \left\{\begin{matrix} a\frac{\left ( r^{n}-1 \right )}{r-1}, &if \: r\neq 1 \\ n\, a, & if \, r= 1 \end{matrix}\right.$

- wherein

$a\rightarrow$ first term

$r\rightarrow$ common ratio

$n\rightarrow$ number of terms

and

(10)9 + 2(11)1 (10)8 + 3(11)2 (10)7 +......  +10 (11)9 = k(10)9

Take common 109

$10^{9}\left [ 1+2\times \frac{11}{10}+3\times \left ( \frac{11}{10} \right )^{2}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot 10\times \left ( \frac{11}{10} \right ) ^{9}\right ]= k\left ( 10 \right )^{9}$

$\therefore k= 1+2x+3x^{2}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot 10x^{9} \: where \:\:x=\frac{11}{10}$

$kx= x+2x^{2}+3x^{3}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot 10x^{10}$

Subtract

$k-kx= 1+x+x^{2}+x^{3}+x^{4}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (-10x^{10})$

$k(1-x)= \frac{1(x^{10}-1)}{x-1}-10x^{10}$

$k\left ( 1-\frac{11}{10} \right )= \frac{\left ( \frac{11}{10} \right )^{10}-1}{\frac{11}{10}-1}-10\times \left ( \frac{11}{10} \right )^{10}$

$-\frac{k}{10}= \frac{\left ( \frac{11}{10} \right )^{10}}{\frac{1}{10}}-10-\frac{11^{10}}{10^{9}}$

$\therefore k=100$

Option 1)

100

Option 2)

110

Option 3)

Option 4)

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Engineering
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If three positive numbers a, b and c are in A.P. such that abc=8, then the minimum possible value of b is :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

Use the Concept of

Arithmetic mean of two numbers (AM) -

$A=\frac{a+b}{2}$

- wherein

It is to be noted that the sequence a, A, b, is in AP where, a and b are the two numbers.

a,b,c in A.P so that

$b=\frac{a+c}{2}$

$2b=a+c$

$\therefore\: c=2b-a$

=> $abc=8$

=>  $ab\left [ 2b-a \right ]=8$

$\therefore \: 2ab^{2} - a^{2}b=8=k$

$\therefore\: 2+1\times b^{2}-2a\times b=0$

$\therefore\: 2b\left [ b-a \right ]=0$

$b\neq 0$

but  $b=a$

then  $c=a$

$\therefore \: a^{3} =8$

$a=2=b=c$

$\therefore \: a=b=c=8$

Option 1)

Correct option

Option 2)

Incorrect option

Option 3)

Incorrect option

Option 4)

Incorrect option

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