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Let $a_{n}$ be the n^th 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 S_n 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

n^th 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, S_n 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 $\Rightarrow$ $a_1 + a_1r = 4$ ………(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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Get Answers to all your Questions

All Questions

Let be the nth term of a G.P. of positive terms. If is equal to : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

Nalla mahalakshmi

If for then Option: 1 Option: 2 Option: 3 Option: 4

Posted by

avinash.dongre

Crack CUET with india's "Best Teachers"

The product is equal to : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

Soham Rahate

The sum is Option: 1 1540 Option: 2 1680 Option: 3 1260 Option: 4 1450

Posted by

Kuldeep Maurya

JEE Main high-scoring chapters and topics

Let be such that for all and are in A.P., then the minimum value of is : Option: 1 Option: 2 Option: 3 Option: 4

Let $a_{n}$ be the n^th 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 $150$

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 $x(1+y)=1$

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 $1$

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

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 $2$

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 $\frac{21}{2}$

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 $63$

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 $1008$

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 $100\frac{1}{2}$

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 $-513$