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Let a_{n} be the nth term of a G.P. of positive terms. If \sum_{n=1}^{100}a_{2n+1}=200\: \: and\: \: \sum_{n=1}^{100}a_{2n}=100,\: \: then\: \sum_{n=1}^{200}a_n is equal to :   
Option: 1 300
Option: 2 175
Option: 3 225
Option: 4 150
 

150

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Posted by

Nalla mahalakshmi

If x=\sum_{n=0}^{\infty }(-1)^{n}\tan ^{2n}\theta \: \: and\: \: y=\sum_{n=0}^{\infty }\cos ^{2n}\theta , for 0<\theta < \frac{\pi }{4}, then
Option: 1 y(1+x)=1
Option: 2 x(1-y)=1
Option: 3 y(1-x)=1
Option: 4 x(1+y)=1
 

x=\sum_{n=0}^{\infty}(-1)^{n} \tan ^{2 n} \theta=1-\tan^2\theta+\tan^4\theta..........

y=\sum_{n=0}^{\infty} \cos ^{2 n} \theta=1+\cos^2\theta+\cos^4\theta......

Use \text S_{\infty}=\frac{1}{1-r}

{x=\frac{1}{1+\tan ^{2} \alpha}=\cos ^{2} \theta} \\ {y=\frac{1}{1-\cos ^{2} \theta}=\frac{1}{\sin ^{2} \theta}}

\Rightarrow (1-x)= \sin ^{2} \theta

\Rightarrow y(1-x)=1

Correct Option (3)

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Posted by

avinash.dongre

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The product  2^{1/4}\cdot 4^{1/16}\cdot 8^{1/48}\cdot 16^{1/128}\cdot \cdots is equal to :   
Option: 1 2^{1/4}
Option: 2 2
Option: 3 2^{1/2}
Option: 4 1
 

Option no 2

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Posted by

Soham Rahate

The sum \sum_{k=1}^{20}(1+2+3+.....+k) is
Option: 1 1540
Option: 2 1680
Option: 3 1260
Option: 4 1450
 

Sum of Common Series

Sum of the square of first n-natural numbers

            \\\mathrm{1^2+2^2+3^2+4^2+...........+n^2=\frac{n(n+1)(2n+1)}{6}}

 

Now,

\sum_{k=1}^{20} \frac{k(k+1)}{2}=\frac{1}{2}\left(\sum_{k=1}^{20} k^{2}+\sum_{k=1}^{20} k\right)

=\frac{1}{2}\left(\frac{20(20+1)(2 \times 20+1)}{6}+\frac{20(20+1)}{2}\right)

=\frac{1}{2}((70 \times 41)+210)=1540

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Posted by

Kuldeep Maurya

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Let f:\; R\rightarrow R be such that for all x\epsilon R (2^{1+x}+2^{1-x}), f(x) and (3^{x}+3^{-x}) are in A.P., then the minimum value of f(x) is :
 
Option: 1 0
Option: 2 4
Option: 3 3
Option: 4 2
 

Important Properties of an AP

If a, b, c are in AP, then 2b = a + c

and 

AM ≥ GM

 

Applying this to given question,

\\2 f(x)=\left(2 \frac{1}{2^{x}}+2.2^{x}\right)+\left(3^{x}+\frac{1}{3^{x}}\right)\\ \\Now,\,\,AM \geq GM\\ \\\frac{a+b}{2}\geq\sqrt{ab}\\\frac{3^{x}+\frac{1}{3^{x}}}{2} \geq \sqrt{1}\\\left(3^{x}+\frac{1}{3^{x}}\right) \geqslant 2\\Value\,\,equals\,\,2 \,\,at\;\;3^x=\frac{1}{3^x}\Rightarrow x=0

\\Similarly, \frac{1}{2^{x}}+2^{x} \geq 2, and\,\,this\,\,holds\,\,at\,\, x=0\\ \\So,2 (minimum\,\,of\,\,f(x))=2(2)+(2)\\\min\;f(x)=3

Correct Option (3)

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Posted by

Kuldeep Maurya

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 :  
Option: 1 16
Option: 2 27
Option: 3 7
Option: 4 \frac{21}{2}
 

Five terms are in AP

Let terms are a-2d, a-d, a, a+d, a+2d

Sum of the terms = 5a = 25

So, a=5

Product of the terms 

(a-2d) (a-d) (a+d) (a+2d) a = 2520

Putting a = 5

(25 - 4d2) (25 - d2) = 504

Putting d2 = t

(25 - t) (25 - 4t) = 504

4t2 - 125t + 121 = 0

t = 1 or t = 121/4

d = 1 or d = -1 or d= 11/2 or d = -11/2

But -1/2 can be a term of AP only for d= 11/2 or d = -11/2

Using d = 11/2

Greatest term = a+ 2d = 16

Correct Option (1)

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Posted by

Kuldeep Maurya

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The greatest positive integer k, for which 49^{k}+1 is a factor of the sum 49^{125}+49^{124}+....+49^{2}+49+1, is :
Option: 1 32
Option: 2 60
Option: 3 65
Option: 4 63

Sum of n-term of a GP

 

Let Sn be the sum of n terms of the G.P. with the first term ‘a’ and common ratio ‘r’. Then 

{S_n=a\left ( \frac{r^n-1}{r-1} \right )}

The above formula does not hold for r = 1, For r = 1 the sum of n terms of the G.P. is S_n = na.

 

Now,

\begin{align*} &1+49+49^2+......49^{125}\\ &\\\ &\Rightarrow \frac{1(49^{126}-1)}{48}\\ &\Rightarrow \frac{(49^{63}-1)(49^{63}+1)}{48}\\ \end{align*}

As xn - yn is always divisible by (x - y), so \frac{(49^{63}-1)}{48} is an integrer

Hence, K=63

Correct Option (4)

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Posted by

Ritika Jonwal

The sum, \sum_{n=1}^{7}\frac{n(n+1)(2n+1)}{4} is equal to  _______.
Option: 1 1024  
Option: 3 810
Option: 5 504
Option: 7 1008
 

Sum of Common Series

Sum of the square of first n-natural numbers

            \\\mathrm{1^2+2^2+3^2+4^2+...........+n^2=\frac{n(n+1)(2n+1)}{6}}Sum of the cube of first n-natural numbers

           \\\mathrm{1^3+2^3+3^3+4^3+...........+n^3=\left \{ \frac{n(n+1)}{2} \right \}^2}

-

Now,

\begin{aligned} &\frac{1}{4}\left[\sum_{n=1}^{7}\left(2 n^{3}+3 n^{2}+n\right)\right]\\ &\frac{1}{4}\left[2\left(\frac{7.8}{2}\right)^{2}+3\left(\frac{7.8 .15}{6}\right)+\frac{7.8}{2}\right]\\ &\frac{1}{4}[2 \times 49 \times 16+28 \times 15+28]\\ &\frac{1}{4}[1568+420+28]=504 \end{aligned}

Correct Option (3)

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Posted by

vishal kumar

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If the 10th term of an A.P. is  \frac{1}{20}  and its 20th term is \frac{1}{10}, then the sum of its first 200 term is :
Option: 1 50\frac{1}{4}
Option: 3 100
Option: 5 50
Option: 7 100\frac{1}{2}
 

General Term of an AP

nth term (general term) of the A.P. is \mathrm{\mathit{a_n=a+(n-1)d}}.

Sum of n terms of an AP

The sum, Sn  of n terms of an AP with the first term ‘a’ and common difference ‘d’ is given by  

 

\begin{array}{l}{S_{n}=\frac{n}{2}[2 a+(n-1) d]} \\ {\text { OR }} \\ {S_{n}=\frac{n}{2}[a+l]} \\ {a \rightarrow \text { first term }} \\ {d \rightarrow \text { common difference }} \\ {n \rightarrow \text { number of terms }}\end{array}

 

Now,

\\a_{10}=a+9d=1/20\\a_{20}=a+19d=1/10\\On\,\,subtracting\,\,them\,\,,we\,\,get\,\,d=1/200\\a+9/10=1/20\Rightarrow a=1/200\\S_{200}=\frac {200}{2}\left [2/200+(199)\times1/200 \right ]=\frac{201}{2}

Correct Option (4)

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Posted by

vishal kumar

Let a_{1},a_{2},a_{3},... be a G.P. such that a_{1}< 0,a_{1}+a_{2}=4 and a_{3}+a_{4}=16. If \sum_{i=1}^{9}a_{i}=4\lambda , then \lambda is equal to :
Option: 1 171


Option: 2 \frac{511}{3}


Option: 3 -171


Option: 4 -513
 

Given,

a1 + a2 = 4 \Rightarrowa_1 + a_1r = 4 ………(i)

a3 + a4 = 16 \Rightarrow a_1r^2 + a_1r ^3= 16 …….(ii)

On dividing these

\begin{array}{l}{\frac{1}{r^{2}}=\frac{1}{4} \Rightarrow r^{2}=4} \\\\ {\Rightarrow r=\pm 2} \\\\ {If\,\, r=2,\,\,then\,\,a_{1}(1+2)=4 \Rightarrow a_{1}=\frac{4}{3}}\end{array}

\begin{array}{l}{If\,\,r=-2, \quad a_{1}(1-2)=4 \Rightarrow a_{1}=-4} \\\\ As\, first\, term \,is \,negative,\, so,\,\, r = -2 \\\\ {\sum_{i=1}^{9} a_{i}=\frac{a_{1}\left(r^{9}-1\right)}{r-1}=\frac{(-4)\left((-2)^{9}-1\right)}{-2-1}=\frac{4}{3}(-513)=4 \lambda} \\ {\lambda=-171}\end{array}

Correct Option (3)

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Posted by

Ritika Jonwal

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