2012
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2012 Paper 2 Offline
Let ${a_1},{a_2},{a_3},.....$ be in harmonic progression with ${a_1} = 5$ and ${a_{20}} = 25.$ The least positive integer $n$ for which ${a_n} < 0$ is
Show Answer
Correct Answer: D
Explanation:
Given: $a_1=5$ and $a_{20}=25$
Also given, $a_1, a_2, a_3, \ldots \ldots \ldots$ are in H.P.
$\Rightarrow \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots \ldots \ldots$ are in A.P.
Let D be the common difference of above A. P.
$\begin{array}{ll}
\therefore & \frac{1}{a_{20}}=\frac{1}{a_1}+(20-1) d \\
\Rightarrow & \frac{1}{25}=\frac{1}{5}+19 d \\
\Rightarrow & d=\frac{-4}{475}
\end{array}$
$\begin{aligned}
& \text { Now, } \quad \frac{1}{a_n}=\frac{1}{a_1}+(n-1) d \\
& \Rightarrow \quad \frac{1}{a_n}=\frac{1}{5}+(n-1) \cdot\left(\frac{-4}{475}\right) \\
& \Rightarrow \quad \frac{1}{a_n}=\frac{95-4 n+4}{475} \\
& \Rightarrow \quad a_n=\frac{475}{99-4 n} \\
\end{aligned}$
Apply $\quad a_n<0$
$\Rightarrow \quad \frac{475}{99-4 n}<0$
The least positive integral value of $n$ is 25 which satisfy the above condition.
2009
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2009 Paper 2 Offline
If the sum of first $n$ terms of an A.P. is $c{n^2}$, then the sum of squares of these $n$ terms is
A.
${{n\left( {4{n^2} - 1} \right){c^2}} \over 6}$
B.
${{n\left( {4{n^2} + 1} \right){c^2}} \over 3}$
C.
${{n\left( {4{n^2} - 1} \right){c^2}} \over 3}$
D.
${{n\left( {4{n^2} + 1} \right){c^2}} \over 6}$
Show Answer
Correct Answer: C
Explanation:
We have ${t_n} = c\{ {n^2} - {(n - 1)^2}\} $
$ = c(2n - 1)$
$ \Rightarrow t_n^2 = {c^2}(4{n^2} - 4n + 1)$
$ \Rightarrow \sum\limits_{n = 1}^n {t_n^2 = {c^2}\left\{ {{{4n(n + 1)(2n + 1)} \over 6} - {{4n(n + 1)} \over 2} + n} \right\}} $
$ = {{{c^2}n} \over 6}\{ 4(n + 1)(2n + 1) - 12(n + 1) + 6\} $
$ = {{{c^2}n} \over 3}\{ 4{n^2} + 6n + 2 - 6n - 6 + 3\} = {{{c^2}} \over 3}n(4{n^2} - 1)$
which is the sum of the square of $n$ terms.
2008
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2008 Paper 2 Offline
Suppose four distinct positive numbers ${a_1},\,{a_{2\,}},\,{a_3},\,{a_4}\,$ are in G.P. Let ${b_1} = {a_1},{b_2} = {b_1} + {a_2},\,{b_3} = {b_2} + {a_{3\,\,}}\,\,\,and\,\,\,{b_4} = {b_3} + {a_4}$.
STATEMENT-1: The numbers ${b_1},\,{b_{2\,}},\,{b_3},\,{b_4}\,$ are neither in A.P. nor in G.P. and
STATEMENT-2 The numbers ${b_1},\,{b_{2\,}},\,{b_3},\,{b_4}\,$ are in H.P.
A.
B.
C.
STATEMENT-1 is True, STATEMENT-2 is False
D.
STATEMENT-1 is False, STATEMENT-2 is True
Show Answer
Correct Answer: C
Explanation:
Given, $a_1,a_2,a_3,a_4$ are in G.P.
Then, $b_1,b_2,b_3,b_4$ are the numbers.
$a_1,a_1+a_2,a_1+a_2+a_3,a_1+a_2+a_3+a_4$ or $a,a+ar,a+ar+ar^2,a+ar+ar^2+ar^3$
Clearly above numbers are neither in A.P. nor in G.P. and hence statement 1 is true.
Also, ${1 \over a},{1 \over {a + ar}},{1 \over {a + ar + a{r^2}}},{1 \over {a + ar + a{r^2} + a{r^3}}}$ are not in H.P.
$\therefore$ $b_1,b_2,b_3,b_4$ are not in H.P.
$\therefore$ Statement 2 is false.
2007
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2007 Paper 2 Offline
Which one of the following statements is correct?
A.
$\mathrm{G}_{1} > \mathrm{G}_{2} > \mathrm{G}_{3} >\ldots$
B.
$\mathrm{G_{1} < G_{2} < G_{3} < \ldots}$
C.
$\mathrm{G}_{1}=\mathrm{G}_{2}=\mathrm{G}_{3}=\ldots$
D.
$\mathrm{G}_{1} < \mathrm{G}_{3} < \mathrm{G}_{5}<\ldots$ and $\mathrm{G}_{2} > \mathrm{G}_{4} > \mathrm{G}_{6} > \ldots$
Show Answer
Correct Answer: C
Explanation:
If A, G, H be A.M, G.M and H.M then $\mathrm{G^{2}=A H}$ and $\mathrm{A} > \mathrm{G} > \mathrm{H}$
Also, we are given
$\mathrm{A}_{n-1}$ and $\mathrm{H}_{n-1}$ have A.M, G.M and H.M as $\mathrm{A}_{n}$, $\mathrm{G}_{n}$ and $\mathrm{H}_{n}$
$\therefore$ By definition, we have
$\mathrm{A}_{n}=\frac{\mathrm{A}_{n-1}+\mathrm{H}_{n-1}}{2}, \mathrm{G}_{n}^{2}=\mathrm{A}_{n-1} \text { and } \mathrm{H}_{n-1}$
$\frac{2}{\mathrm{H}_{n}}=\frac{1}{\mathrm{~A}_{n-1}}+\frac{1}{\mathrm{H}_{n-1}} \text { for }$
$n=2, \mathrm{G}_{2}^{2}=\mathrm{A}_{1} \mathrm{H}_{1}=\mathrm{G}_{1}^{2} \text { so on }
$
$\therefore G_{1}^{2}=G_{2}^{2}=G_{3}^{2} \ldots \ldots$
$\Rightarrow \mathrm{G}_{1}=\mathrm{G}_{2}=\mathrm{G}_{3} \ldots \ldots$
2007
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2007 Paper 2 Offline
Which one of the following statements is correct?
A.
$A_{1} > A_{2} > A_{3} > \ldots$
B.
$\mathrm{A}_{1} < \mathrm{A}_{2} < \mathrm{A}_{3} < \ldots$
C.
$A_{1} > A_{3} > A_{5}>\ldots$ and $A_{2} < A_{4} < A_{6} < \ldots$
D.
$A_{1} < A_{3} < A_{5} < \ldots$ and $A_{2}>A_{4} > A_{6} > \ldots$
Show Answer
Correct Answer: A
Explanation:
$A_{2}$ is A.M. of $A_{1}, H_{1}$ and $A_{1} > H$
$\Rightarrow \mathrm{A}_{1} > \mathrm{A}_{2} > \mathrm{H}_{1}$
$\mathrm{A}_{3}$ is A.M. of $\mathrm{A}_{2}, \mathrm{H}_{2}$
$\mathrm{A}_{2} > \mathrm{A}_{3} > \mathrm{H}_{2}$
$\therefore \mathrm{A}_{1} > \mathrm{A}_{2} > \mathrm{A}_{3} \ldots \ldots$.
2007
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2007 Paper 2 Offline
Which one of the following statements is correct?
A.
$\mathrm{H}_{1} > \mathrm{H}_{2} > \mathrm{H}_{3} > \ldots$
B.
$\mathrm{H}_{1} < \mathrm{H}_{2} < \mathrm{H}_{3} < \ldots$
C.
$\mathrm{H}_{1}>\mathrm{H}_{3} > \mathrm{H}_{5} > \ldots$ and $\mathrm{H}_{2} < \mathrm{H}_{4} < \mathrm{H}_{6} < \ldots$
D.
$\mathrm{H}_{1} < \mathrm{H}_{3} < \mathrm{H}_{5}< \ldots$ and $\mathrm{H}_{2} > \mathrm{H}_{4} > \mathrm{H}_{6} > \ldots$
Show Answer
Correct Answer: B
Explanation:
let $a$ and $b$ are two numbers then
$\begin{aligned}& \mathrm{A}_{1}=\frac{a+b}{2} ; \mathrm{G}_{1}=\sqrt{a b}, \mathrm{H}_{1}=\frac{2 a b}{a+b} \\
& \mathrm{~A}_{n}=\frac{\mathrm{A}_{n-1}+\mathrm{H}_{n-1}}{2}
\end{aligned}$
$\mathrm{G}_{n}=\frac{2 \mathrm{~A}_{n-1}+\mathrm{H}_{n-1}}{\mathrm{~A}_{n-1}+\mathrm{H}_{n-1}}$
$G_{1}=G_{2}=G_{3}=\ldots . \cdot \sqrt{a b}$
$A_{2}$ is A.M. of $A_{1}$ and $H_{1}$ and
$\mathrm{A}_{1} > \mathrm{H}_{1} \Rightarrow \mathrm{A}_{1} > \mathrm{A}_{2} > \mathrm{H}_{1}$
$A_{3}$ is A.M. of $A_{2}$ and $H_{2}$
$\mathrm{A}_{2} > \mathrm{H}_{2} \Rightarrow \mathrm{A}_{2} > \mathrm{A}_{3} > \mathrm{H}_{2}$ .......
$\therefore \mathrm{A}_{1} > \mathrm{A}_{2} > \mathrm{A}_{3} > \ldots$
$\mathrm{A}_{1} > \mathrm{H}_{2} > \mathrm{H}_{1} > \mathrm{A}_{2} > \mathrm{H}_{3} > \mathrm{H}_{2}$
$\therefore \mathrm{H}_{1} < \mathrm{H}_{2} < \mathrm{H}_{3} \ldots \ldots$.
2007
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2007 Paper 1 Offline
The sum V$_1$ + V$_2$ + ... + V$_n$ is
A.
${1 \over {12}}n(n + 1)(3{n^2} - n + 1)$
B.
${1 \over {12}}n(n + 1)(3{n^2} + n + 2)$
C.
${1 \over 2}n(2{n^2} - n + 1)$
D.
${1 \over 3}(2{n^3} - 2n + 3)$
Show Answer
Correct Answer: B
Explanation:
$\sum\limits_{r = 1}^n {{v_r} = \sum\limits_{r = 1}^n {\left( {{r \over 2}(2r + (r - 1)(2r - 1))} \right)} } $
$ = \sum\limits_{r = 1}^n {\left( {{r^3} - {{{r^2}} \over 2} + {r \over 2}} \right)} $
$ = \sum {{n^3} - {1 \over 2}\sum {{n^2} + {1 \over 2}\sum n } } $
$ = {{{n^2}{{(n + 1)}^2}} \over 4} - {{n(n + 1)(2n + 1)} \over {12}} + {{n(n + 1)} \over {2 \times 2}}$
$ = {{n(n + 1)} \over 4}\left[ {n(n + 1) - {{(2n + 1)} \over 3} + 1} \right]$
$ = {{n(n + 1)[3{n^2} + n + 2]} \over {12}}$
$ = {1 \over {12}}n(n + 1)[3{n^2} + n + 2]$
2007
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2007 Paper 1 Offline
Show Answer
Correct Answer: D
Explanation:
We have,
${v_r} = {1 \over 2}(2{r^3} - {r^2} + r)$
${v_{r + 1}} = {1 \over 2}[2{(r + 1)^3} - {(r + 1)^2} + (r + 1)]$
Now,
${T_r} = {v_{r + 1}} - {v_r} - 2$
$ = [{(r + 1)^3} - {r^3}] - {1 \over 2}[{(r + 1)^2} - {r^2}] + {1 \over 2}(1) - 2$
$ = 3{r^2} + 2r - 1$
$ = (r + 1)(3r - 1)$
Which is a composite number
2007
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2007 Paper 1 Offline
Which one of the following is a correct statement?
A.
Q$_1$, Q$_2$, Q$_3$, ... are in A.P. with common difference 5
B.
Q$_1$, Q$_2$, Q$_3$, ... are in A.P. with common difference 6
C.
Q$_1$, Q$_2$, Q$_3$, ... are in A.P. with common difference 11
D.
Q$_1$ = Q$_2$ = Q$_3$, ...
Show Answer
Correct Answer: B
Explanation:
We have
${T_r} = 3{r^2} + 2r - 1$
${T_{r + 1}} = 3{(r + 1)^2} + 2(r + 1) - 1$
Now,
${Q_r} = {T_{r + 1}} - {T_r} = 3[{(r + 1)^2} - {r^2}] + 2(1)$
$ = 6r + 5$
${Q_{r + 1}} = 6(r + 1) + 5$
${Q_{r + 1}} - {Q_r} = 6 = $ constant
Therefore, Q$_1$, Q$_2$ and Q$_3$ ..... are in A.P. with common difference 6.
2005
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2005 Screening
In the quadratic equation $\,\,a{x^2} + bx + c = 0,$ $\Delta $ $ = {b^2} - 4ac$ and $\alpha + \beta ,\,{\alpha ^2} + {\beta ^2},\,{\alpha ^3} + {\beta ^3},$ are in G.P. where $\alpha ,\beta $ are the root of $\,\,a{x^2} + bx + c = 0,$ then
Show Answer
2005
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2005 Mains
If total number of runs scored in $n$ matches is $\left(\frac{n+1}{4}\right)\left(2^{n+1}-n-2\right)$ where $n > 1$, and the runs scored in the $k^{\text {th }}$ match are given by $k .2^{n+1-k}$, where $1 \leq k \leq n$. Find, $n$.
Show Answer
Correct Answer: B
Explanation:
let $S_{n}$ be the sum of all.
run scored in $\mathrm{K}$ - matches.
i.e., $\mathrm{S}_{\mathrm{n}}=\sum_\limits{k=1}^{n} k .2^{n+1-k}$
$\begin{aligned}& =2^{n+1} \sum_{k=1}^{n} k \cdot 2^{-k} \\
S_{n} & =2^{n+1}\left[\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\ldots . \cdot \frac{n}{2^{n}}\right]\quad .... \mathrm{(i)}
\end{aligned}$
Multiply $S_{n}$ by $\frac{1}{2}$, we have.
$\frac{S_{n}}{2}=2^{n+1}\left[\frac{1}{2^{2}}+\frac{2}{2^{3}}+\frac{3}{2^{4}}+\ldots \ldots+\frac{n}{2^{n+1}}\right]$ ..... (ii)
Subtract (ii) from (i), we have.
$\begin{aligned}
& \frac{S_{n}}{2}=2^{n+1}\left[\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots .+\frac{1}{2^{n}}-\frac{n}{2^{n+1}}\right] \\
&=2^{n+1}\left[\frac{\frac{1}{2}\left(1-\frac{1}{2^{n}}\right)}{\frac{1}{2}}-\frac{n}{2^{n+1}}\right] \\
&= 2^{n+1}\left[\frac{2^{n}-1}{2^{n}}-\frac{n}{2^{n+1}}\right] \\
&= {\left[2^{n+1}-2-n\right] } \\
& \Rightarrow \quad S_{n}= 2\left[2^{n+1}-2-n\right] \\
& \text { Now, } S_{n}= \frac{n+1}{4}\left(2^{n+1}-n-2\right) \text { given, } \\ & 2({2^{n + 1}} - 2 - n) = {{n + 1} \over 4}({2^{n + 1}} - n - 2) \\
& \Rightarrow \quad \frac{n+1}{4}=2\end{aligned}$
$\Rightarrow \quad n+1=8\quad \Rightarrow n=7$
2004
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2004 Screening
An infinite G.P. has first term '$x$' and sum '$5$', then $x$ belongs to
Show Answer
2002
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2002 Screening
Suppose $a, b, c$ are in A.P. and ${a^2},{b^2},{c^2}$ are in G.P. If $a < b < c$ and $a + b + c = {3 \over 2},$ then the value of $a$ is
A.
${1 \over {2\sqrt 2 }}$
B.
${1 \over {2\sqrt 3 }}$
C.
${1 \over 2} - {1 \over {\sqrt 3 }}$
D.
${1 \over 2} - {1 \over {\sqrt 2 }}$
Show Answer
2001
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2001 Screening
Let the positive numbers $a,b,c,d$ be in A.P. Then $abc,$ $abd,$ $acd,$ $bcd,$ are
Show Answer
2001
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2001 Screening
If the sum of the first $2n$ terms of the A.P.$2,5,8,......,$ is equal to the sum of the first $n$ terms of the A.P.$57,59,61,.....,$ then $n$ equals
Show Answer
2001
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2001 Screening
Let $\alpha $, $\beta $ be the roots of ${x^2} - x + p = 0$ and $\gamma ,\delta $ be the roots of ${x^2} - 4x + q = 0.$ If $\alpha ,\beta ,\gamma ,\delta $ are in G.P., then the integral values of $p$ and $q$ respectively, are
Show Answer
2000
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2000 Screening
Consider an infinite geometric series with first term a and common ratio $r$. If its sum is 4 and the second term is 3/4, then
A.
$a = {4 \over 7},r = {3 \over 7}\,\,\,\,$
B.
$a = 2,\,r = {3 \over 8}$
C.
$a = {3 \over 2},r = {1 \over 2}$
D.
$a = 3,\,r = {1 \over 4}$
Show Answer
1999
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1999
The harmonic mean of the roots of the equation $\left( {5 + \sqrt 2 } \right){x^2} - \left( {4 + \sqrt 5 } \right)x + 8 + 2\sqrt 5 = 0$ is
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1999
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1999
Let ${a_1},{a_2},......{a_{10}}$ be in $A,\,P,$ and ${h_1},{h_2},......{h_{10}}$ be in H.P. If ${a_1} = {h_1} = 2$ and ${a_{10}} = {h_{10}} = 3,$ then ${a_4}{h_7}$ is
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1998
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1998
Let $n$ be an odd integer. If $\sin n\theta = \sum\limits_{r = 0}^n {{b_r}{{\sin }^r}\theta ,} $ for every value of $\theta ,$ then
B.
${b_0} = 0,\,{b_1} = n$
C.
${b_0} = - 1,\,{b_1} = n$
D.
${b_0} = 0,\,{b_1} = {n^2} - 3n + 3$
Show Answer
1998
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1998
If $x > 1,y > 1,z > 1$ are in G.P., then ${1 \over {1 + In\,x}},{1 \over {1 + In\,y}},{1 \over {1 + In\,z}}$ are in
Show Answer
1998
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1998
Let ${T_r}$ be the ${r^{th}}$ term of an A.P., for $r=1, 2, 3, ....$ If for some positive integers $m$, $n$ we have
B.
${1 \over {mn}} + {1 \over n}$
Show Answer
1994
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1994
If $In\left( {a + c} \right),In\left( {a - c} \right),In\left( {a - 2b + c} \right)$ are in A.P., then
A.
$a,\,b,\,c$ are in A.P.
B.
${a^2},\,{b^2},\,{c^2}$ are in A.P.
C.
$a,\,b,\,c$ are in G.P.
D.
$a,\,b,\,c$ are in H.P.
Show Answer
1990
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1990
The number ${\log _2}\,7$ is
Show Answer
1988
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1988
Sum of the first n terms of the series ${1 \over 2} + {3 \over 4} + {7 \over 8} + {{15} \over {16}} + ............$ is equal to
Show Answer
1985
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1985
If $a,\,b,\,c$ are in GP., then the equations $\,\,\alpha {x^2} + 2bx + c = 0$ and $d{x^2} + 2ex + f = 0$ have a common root if ${d \over a},\,{e \over b},{f \over c}$ are in ________.
Show Answer
1983
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1983
The rational number, which equals the number $2\overline {357} $ with recurring decimal is
A.
${{2355} \over {1001}}$
B.
${{2379} \over {997}}$
C.
${{2355} \over {999}}$
Show Answer
1982
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1982
The third term of a geometric progression is 4. The product of the first five terms is
Show Answer
1982
JEE Advanced
MCQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1982
If $x,\,y$ and $z$ are $pth$, $qth$ and $rth$ terms respectively of an A.P. and also of a G.P., then ${x^{y - z}}\,{y^{z - x}}\,{z^{x - y}}$ is equal to :
Show Answer
2023
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2023 Paper 1 Online
Let $7 \overbrace{5 \cdots 5}^r 7$ denote the $(r+2)$ digit number where the first and the last digits are 7 and the remaining $r$ digits are 5 . Consider the sum $S=77+757+7557+\cdots+7 \overbrace{5 \cdots 5}^{98}7$. If $S=\frac{7 \overbrace{5 \cdots 5}^{99}7+m}{n}$, where $m$ and $n$ are natural numbers less than 3000 , then the value of $m+n$ is
Show Answer
Correct Answer: 1219
Explanation:
$S=77+757+7557+\cdots+ 7\overbrace{5 \cdots 5}^{98}7$
2022
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2022 Paper 1 Online
Let $l_{1}, l_{2}, \ldots, l_{100}$ be consecutive terms of an arithmetic progression with common difference $d_{1}$, and let $w_{1}, w_{2}, \ldots, w_{100}$ be consecutive terms of another arithmetic progression with common difference $d_{2}$, where $d_{1} d_{2}=10$. For each $i=1,2, \ldots, 100$, let $R_{i}$ be a rectangle with length $l_{i}$, width $w_{i}$ and area $A_{i}$. If $A_{51}-A_{50}=1000$, then the value of $A_{100}-A_{90}$ is __________.
Show Answer
Correct Answer: 18900
Explanation:
Given,
${l_1},{l_2},\,.......,\,{l_{100}}$ are in A.P with common difference ${d_1}$.
So from property of A.P we can say,
${l_2} = {l_1} + {d_1}$
${l_3} = {l_1} + 2{d_1}$
$ \vdots $
${l_{100}} = {l_1} + 99{d_1}$
Also given,
${w_1},{w_2},\,......\,,\,{w_{100}}$ are in A.P with common difference ${d_2}$.
$\therefore$ From the property of A.P we can say,
${w_2} = {w_1} + {d_2}$
${w_3} = {w_1} + 2{d_2}$
$ \vdots $
${w_{100}} = {w_1} + 99{d_2}$
Now, also given,
${d_1}{d_2} = 10$
and ${R_i}$ is a rectangle whose length is ${l_i}$ and width is ${w_i}$ and area ${A_i}$.
$\therefore$ We know, area of rectangle
${A_i} = {l_i} \times {w_i}$
$\therefore$ ${A_{51}} = {l_{51}} \times {w_{51}}$
and ${A_{50}} = {l_{50}} \times {w_{50}}$
Given, ${A_{51}} - {A_{50}} = 1000$
$ \Rightarrow ({l_1} + 50{d_1})({w_1} + 50{d_2}) - ({l_1} + 49{d_1})({w_1} + 49{d_2}) = 1000$
$ \Rightarrow [{l_1}{w_1} + 2500{d_1}{d_2} + 50({l_1}{d_2} + {d_1}{w_1})] - [{l_1}{w_1} + 49 \times 49{d_1}{d_2} + 49({l_1}{d_2} + {w_1}{d_1})] = 1000$
$ \Rightarrow [{(50)^2}{d_1}{d_2} - {(49)^2}{d_1}{d_2}] + (50 - 49)({l_1}{d_2} + {d_1}{w_1}) = 1000$
$ \Rightarrow (99 \times 1){d_1}{d_2} + {l_1}{d_2} + {d_1}{w_1} = 1000$
$ \Rightarrow 99 \times 10 + {l_1}{d_2} + {w_1}{d_1} = 1000$
$ \Rightarrow {l_1}{d_2} + {w_1}{d_1} = 10$
Now,
${A_{100}} - {A_{90}}$
$ = {l_{100}}\,.\,{w_{100}} - {l_{90}}\,.\,{w_{90}}$
$ = ({l_1} + 99{d_1})({w_1} + 99{d_2}) - ({l_1} + 89{d_1})({w_1} + 89{d_2})$
$ = [{l_1}{w_1} + {(99)^2}{d_1}{d_2} + 99({l_1}{d_2} + {w_1}{d_1})] - [{l_1}{w_1} + {(89)^2}{d_1}{d_2} + 89({l_1}{d_2} + {w_1}{d_1})]$
$ = [{(99)^2} - {89^2}]{d_1}{d_2} + 10({l_1}{d_2} + {w_1}{d_1})$
$ = 188 \times 10 \times {d_1}{d_2} + 10 \times 10$
$ = 188 \times 10 \times 10 + 100$
$ = 18800 + 100$
$ = 18900$
2020
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2020 Paper 1 Offline
Let m be the minimum possible value of ${\log _3}({3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}})$, where ${y_1},{y_2},{y_3}$ are real numbers for which ${{y_1} + {y_2} + {y_3}}$ = 9. Let M be the maximum possible value of $({\log _3}{x_1} + {\log _3}{x_2} + {\log _3}{x_3})$, where ${x_1},{x_2},{x_3}$ are positive real numbers for which ${{x_1} + {x_2} + {x_3}}$ = 9. Then the value of ${\log _2}({m^3}) + {\log _3}({M^2})$ is ...........
Show Answer
Correct Answer: 8
Explanation:
For real numbers y1 , y2 , y3 , the quantities ${{3^{{y_1}}}}$, ${{3^{{y_2}}}}$ and ${{3^{{y_3}}}}$ are positive real numbers, so according to the AM-GM inequality, we have1 , x2 and x3 according to AM-GM inequality, we have1 + x2 + x3 = 9}
2020
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2020 Paper 1 Offline
Let a1 , a2 , a3 , .... be a sequence of positive integers in arithmetic progression with common difference 2. Also, let b1 , b2 , b3 , .... be a sequence of positive integers in geometric progression with common ratio 2. If a1 = b1 = c, then the number of all possible values of c, for which the equality 2(a1 + a2 + ... + an ) = b1 + b2 + ... + bn holds for some positive integer n, is ...........
Show Answer
Correct Answer: 1
Explanation:
Given arithmetic progression of positive integers terms a1 , a2 , a3 , ..... having common difference '2' and geometric progression of positive integers 1 , b2 , b3 , .... having common ratio '2' with a1 = b1 = c, such that 1 + a2 + a3 + ... + an ) = b1 + b2 + b3 + ... + bn
2019
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2019 Paper 1 Offline
Let AP(a; d) denote the set of all the terms of an infinite arithmetic progression with first term a and common difference d > 0. If $AP(1;3) \cap AP(2;5) \cap AP(3;7)$ = AP(a ; d), then a + d equals ..............
Show Answer
Correct Answer: 157
Explanation:
Given that, AP(a ; d) denote the set of all the terms of an infinite arithmetic progression with first term 'a' and common difference d > 0.th term of first progressionth term of progressionth term of third progression
2018
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2018 Paper 1 Offline
Let X be the set consisting of the first 2018 terms of the arithmetic progression 1, 6, 11, ...., and Y be the set consisting of the first 2018 terms of the arithmetic progression 9, 16, 23, .... . Then, the number of elements in the set X $ \cup $ Y is .........
Show Answer
Correct Answer: 3748
Explanation:
Here, X = {1, 6, 11, ....., 10086}n = a + (n $-$ 1)d]n of X $ \cap $ Y is less than or equal to 10086n = 16 + (n $-$ 1) 35 $ \le $ 10086
2017
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2017 Paper 1 Offline
The sides of a right angled triangle are in arithmetic progression. If the triangle has area 24, then what is the length of its smallest side?
Show Answer
Correct Answer: 6
Explanation:
Let the sides be given by a $-$ d, a, a + d, where (a, d > 0). Also, d < a.
By the condition,
a2 + (a $-$ d)2 = (a + d)2
$\Rightarrow$ a2 = (a + d)2 $-$ (a $-$ d)2
$\Rightarrow$ a2 = 4ad $\therefore$ a = 4d
Thus the sides are 3d, 4d, 5d.
As area = 24, we have ${1 \over 2}$ . 3d . 4d = 24
$\therefore$ d = 2
The sides are 6, 8, 10./p>
2015
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2015 Paper 2 Offline
Suppose that all the terms of an arithmetic progression (A.P) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is
Show Answer
Correct Answer: 9
Explanation:
${{{S_7}} \over {{S_{11}}}} = {6 \over {11}}$ ...... (1)
$130 \le {t_7} \le 140$ ........ (2)
$ \Rightarrow {{{7 \over 2}[2a + 6d]} \over {{{11} \over 2}[2a + 10d]}} = {6 \over {11}}$
$ \Rightarrow {{a + 3d} \over {a + 5d}} = {6 \over 7}$ ....... (3)
$ \Rightarrow {{{t_4}} \over {{4_6}}} = {6 \over 7}$
Let ${t_4} = 6k$, ${t_6} = 7k$;
$2d = k \Rightarrow d = k/2$ and $a + 3d = 6k$
$ \Rightarrow a = 6k - 3k/2 = 9k/2$
Hence, $130 \le {t_7} \le 140$.
$ \Rightarrow 130 \le {{9k} \over 2} + 3k \le 140$
$ \Rightarrow 130 \le {{15k} \over 2} \le 140$
$ \Rightarrow {{52} \over 3} \le k \in {{56} \over 3}$
Since, $k \in N \Rightarrow k = 18$.
$ \Rightarrow d = {k \over 2} = {{18} \over 2} = 9$
2015
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2015 Paper 2 Offline
The coefficient of ${x^9}$ in the expansion of (1 + x) (1 + ${x^2)}$ (1 + ${x^3}$) ....$(1 + {x^{100}})$ is
Show Answer
Correct Answer: 8
Explanation:
Given expression is
$E = (1 + x)(1 + {x^2})(1 + {x^3})......(1 + {x^{100}})$
Coefficient of x9 in E
= coefficient of x9 in $(1 + x)(1 + {x^2})(1 + {x^3})......(1 + {x^9})$
$\Rightarrow$ Terms containing x9
$ = (1\,.\,{x^9} + {x^1}\,.\,{x^8} + {x^2}\,.\,{x^7} + {x^3}\,.\,{x^6} + {x^4}\,.\,{x^5} + {x^1}\,.\,{x^2}\,.\,{x^6} + {x^1}\,.\,{x^3}\,.\,{x^5} + {x^2}\,.\,{x^3}\,.\,{x^4})$
$\Rightarrow$ Term containing x9 is 8x9 in E $\Rightarrow$ Coefficient of x9 = 8.
2014
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2014 Paper 1 Offline
Let a, b, c be positive integers such that ${b \over a}$ is an integer. If a, b, c are in geometric progression and the arithmetic mean of a, b, c is b + 2, then the value of ${{{a^2} + a - 14} \over {a + 1}}$ is
Show Answer
Correct Answer: 4
Explanation:
Let $a=a, b=a r$ and $c=a r^2$, where $r$ is integer since ${b \over a}$ is an integer.
2013
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2013 Paper 1 Offline
A pack contains $n$ cards numbered from $1$ to $n.$ Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is $1224.$ If the smaller of the numbers on the removed cards is $k,$ then $k-20=$
Show Answer
Correct Answer: 5
Explanation:
Let number of removed cards are $k$ and $k+1$.
Given, the sum of numbers on the cards after removing $k$ and $k+1$ is 1224.
$\begin{aligned}
& \therefore(1+2+3+\ldots .+n)-(k+(k+1))=1224 \\
& \Rightarrow \quad \frac{n(n+1)}{2}-2 k-1=1224 \\
& \Rightarrow \quad \frac{n(n+1)}{2}-2 k=1225 \\
& \Rightarrow \quad n^2+n-4 k=2450 \\
& \Rightarrow \quad n^2+n-2450=4 k \\
& \Rightarrow(n+50)(n-49)=4 k \\
& \therefore \quad n>49 \\
& \text { Let } n=50 \\
& \Rightarrow \quad 100 \times 1=4 k \\
\end{aligned}$
$\begin{aligned}
\Rightarrow \quad k =25 \\
\Rightarrow \quad k-20 =5
\end{aligned}$
Hints :
(i) Recall $1+2+3+\ldots+n=\frac{n(n+1)}{2}$
(ii) If $k$ is the smallest number on two consecutive numbers, then the second number is $k+1$.
2011
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2011 Paper 1 Offline
Let ${{a_1}}$, ${{a_2}}$, ${{a_3}}$........ ${{a_{100}}}$ be an arithmetic progression with ${{a_1}}$ = 3 and ${S_p} = \sum\limits_{i = 1}^p {{a_i},1 \le } \,p\, \le 100$. For any integer n with $1\,\, \le \,n\, \le 20$, let m = 5n. If ${{{S_m}} \over {{S_n}}}$ does not depend on n, then ${a_{2\,}}$ is
Show Answer
Correct Answer: 9
Explanation:
It is given that a1 , a2 , a3 , ......, a100 is an A.P.
${a_1} = 3,\,{S_p} = \sum\limits_{i = 1}^p {{a_1}} ,\,1 \le p \le 100$
${{{S_m}} \over {{S_n}}} = {{{S_{5n}}} \over {{S_n}}} = {{{{5n} \over 2}(6 + (5n - 1)d)} \over {{n \over 2}(6 - d + nd)}}$
${{{S_m}} \over {{S_n}}}$ is independent of n of $6 - d = 0 \Rightarrow d = 6$.
Therefore, ${a_2} = {a_1} + d = 3 + 6 = 9$.
2010
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2010 Paper 2 Offline
Let ${a_1},\,{a_{2\,}},\,{a_3}$......,${a_{11}}$ be real numbers satisfying ${a_1} = 15,27 - 2{a_2} > 0\,\,and\,\,{a_k} = 2{a_{k - 1}} - {a_{k - 2}}\,\,for\,k = 3,4,........11$. if $\,\,\,{{a_1^2 + a_2^2 + .... + a_{11}^2} \over {11}} = 90$, then the value of ${{{a_1} + {a_2} + .... + {a_{11}}} \over {11}}$ is equal to :
Show Answer
Correct Answer: 0
Explanation:
${a_k} = 2{a_{k - 1}} - {a_{k - 2}}$
$ \Rightarrow {a_1},{a_2},\,\,.....,\,\,{a_{11}}$ are in AP
$\therefore$ ${{a_1^2 + a_2^2 + \,\,....\,\, + \,\,a_{11}^2} \over {11}}$
$ = {{11{a^2} + 35 \times 11{d^2} + 10ad} \over {11}} = 90$
$ \Rightarrow 225 + 35{d^2} + 150d = 90$
$35{d^2} + 150d + 135 = 0$
$ \Rightarrow d = - 3, - {9 \over 7}$
Given, ${a_2} < {{22} \over 7}$ $\therefore$ $d = - 3$ and $d \ne - {9 \over 7}$
$ \Rightarrow {{{a_1} + {a_2} + \,\,...\,\, + \,\,{a_{11}}} \over {11}}$
$ = {{11} \over 2}[30 - 10 \times 3] = 0$
2010
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2010 Paper 1 Offline
Let ${S_k}$= 1, 2,....., 100, denote the sum of the infinite geometric series whose first term is $\,{{k - 1} \over {k\,!}}$ and the common ratio is ${1 \over k}$. Then the value of ${{{{100}^2}} \over {100!}}\,\, + \,\,\sum\limits_{k = 1}^{100} {\left| {({k^2} - 3k + 1)\,\,{S_k}} \right|\,\,} $ is
Show Answer
Correct Answer: 3
Explanation:
$\begin{aligned} & \text { Using } S_{\infty}=\frac{a}{1-r} \text {, we get } \\\\ & \qquad S_k=\left\{\begin{array}{cc}0, & k=1 \\\\ \frac{1}{(k-1)!}, & k \geq 2\end{array}\right.\end{aligned}$
$\begin{aligned} & \text { Now } \sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right|=\sum_{k=2}^{100}\left|\left(k^2-3 k+1\right)\right| \frac{1}{(k-1)!} \\\\ &=|-1|+\sum_{k=3}^{100} \frac{\left(k^2-1\right)+1-3(k-1)-2}{(k-1)!} \\\\ & \quad \text { as } k^2-3 k+1>0 \forall k \geq 3\end{aligned}$
$\begin{aligned} & =1+\sum_{k=3}^{100}\left(\frac{1}{(k-3)!}-\frac{1}{(k-1)!}\right) \\\\ & =1+\left(1-\frac{1}{2!}\right)+\left(\frac{1}{1!}-\frac{1}{3!}\right)+\left(\frac{1}{2!}-\frac{1}{4!}\right)+\ldots+ \\\\ & \qquad\left(\frac{1}{96!}-\frac{1}{98!}\right)+\left(\frac{1}{97!}-\frac{1}{99!}\right)\end{aligned}$
$ \begin{aligned} & =3-\frac{1}{98!}-\frac{1}{99!}=3-\frac{9900}{100!}-\frac{100}{100!} \\\\ & =3-\frac{10000}{100!}=3-\frac{(100)^2}{100!} \\\\ & \therefore \frac{100^2}{100!}+\sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right|=3 \end{aligned} $
1990
JEE Advanced
Numerical
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1990
If ${\log _3}\,2\,,\,\,{\log _3}\,({2^x} - 5)\,,\,and\,\,{\log _3}\,\left( {{2^x} - {7 \over 2}} \right)$ are in arithmetic progression, determine the value of x.
Show Answer
2022
JEE Advanced
MSQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2022 Paper 1 Online
Let $a_{1}, a_{2}, a_{3}, \ldots$ be an arithmetic progression with $a_{1}=7$ and common difference 8. Let $T_{1}, T_{2}, T_{3}, \ldots$ be such that $T_{1}=3$ and $T_{n+1}-T_{n}=a_{n}$ for $n \geq 1$. Then, which of the following is/are TRUE ?
B.
$\sum\limits_{k=1}^{20} T_{k}=10510$
D.
$\sum\limits_{k=1}^{30} T_{k}=35610$
Show Answer
Correct Answer: B,C
Explanation:
Here $a_n=7+(n-1) 8$ and $\mathrm{T}_1=3$, $a_1=7, d=8$
2021
JEE Advanced
MSQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2021 Paper 1 Online
For any positive integer n, let Sn : (0, $\infty$) $\to$ R be defined by ${S_n}(x) = \sum\nolimits_{k = 1}^n {{{\cot }^{ - 1}}\left( {{{1 + k(k + 1){x^2}} \over x}} \right)} $, where for any x $\in$ R, ${\cot ^{ - 1}}(x) \in (0,\pi )$ and ${\tan ^{ - 1}}(x) \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$. Then which of the following statements is (are) TRUE?
A.
${S_{10}}(x) = {\pi \over 2} - {\tan ^{ - 1}}\left( {{{1 + 11{x^2}} \over {10x}}} \right)$, for all x > 0
B.
$\mathop {\lim }\limits_{n \to \infty } \cot ({S_n}(x)) = x$, for all x > 0
C.
The equation ${S_3}(x) = {\pi \over 4}$ has a root in (0, $\infty$)
D.
$tan({S_n}(x)) \le {1 \over 2}$, for all n $\ge$ 1 and x > 0
Show Answer
Correct Answer: A,B
Explanation:
For option (a) ${S_n}(x) = \sum\limits_{k = 1}^n {{{\cot }^{ - 1}}\left[ {{{1 + k(k + 1){x^2}} \over x}} \right]} $ can be written as
2013
JEE Advanced
MSQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Advanced 2013 Paper 1 Offline
Let ${S_n} = {\sum\limits_{k = 1}^{4n} {\left( { - 1} \right)} ^{{{k\left( {k + 1} \right)} \over 2}}}{k^2}.$ Then ${S_n}$can take value(s)
Show Answer
Correct Answer: A,D
Explanation:
$\begin{aligned}
& \text { Given, } \mathrm{S}_n=\sum_{k=1}^{4 n}(-1)^{\frac{k(k+1)}{2}} \cdot \mathrm{K}^2 \\
& \Rightarrow \mathrm{S}_n=-1^2-2^2+3^2+4^2-5^2-6^2+7^2 +8^2-9^2-10^2+11^2+12^2 \ldots \ldots . 4 n \text { terms }
\end{aligned}$
$\begin{aligned}
\Rightarrow \mathrm{S}_n= & -\left[1^2+5^2+9^2+\ldots . n \text { terms }\right] \\
& -\left[2^2+6^2+10^2+\ldots n \text { terms }\right] \\
& +\left[3^2+7^2+11^2+\ldots n \text { terms }\right] \\
& +\left[4^2+8^2+12^2+\ldots n \text { terms }\right]
\end{aligned}$
$\Rightarrow \mathrm{S}_n=-\sum_\limits{r=1}^n(4 r-3)^2-\sum_\limits{r=1}^n(4 r-2)^2 +\sum_\limits{r=1}^n(4 r-1)^2+\sum_\limits{r=1}^n(4 r)^2$
$\begin{aligned}
& \Rightarrow \mathrm{S}_n=\sum_{r=1}^n\left((4 r)^2+(4 r-1)^2-(4 r-2)^2-(4 r-3)^2\right) \\
& \Rightarrow \mathrm{S}_n=\sum_{r=1}^n(32 r-12) \\
& \Rightarrow \mathrm{S}_n=32 \sum_{r=1}^n r-\sum_{r=1}^n 12 \\
& \Rightarrow \mathrm{S}_n=32 \cdot \frac{n(n+1)}{2}-12 n \\
& \Rightarrow \mathrm{S}_n=16 n^2+16 n-12 n \\
& \Rightarrow \mathrm{S}_n=4 n(4 n+1)
\end{aligned}$
If $n=9$, then $\mathrm{S}_9=1332$
If $n=8$, then $\mathrm{S}_8=1056$
Hints :
(i) Recall $\sum_\limits{\mathrm{K}=1}^{4 n}(-1)^{\frac{k(k+1)}{2}} \cdot \mathrm{K}^2=-\sum_\limits{r=1}^n(4 r-3)^2-\sum_\limits{r=1}^n(4 r-2)^2+\sum_\limits{r=1}^n(4 r-1)^2+\sum_\limits{r=1}^n(4 r)^2$
(ii) $\sum_\limits{r=1}^n a \cdot f(r)+b \cdot g(r)-c \cdot h(r)
= \sum_\limits{r=1}^n a \cdot f(r)+\sum_\limits{r=1}^n b \cdot g(r)-\sum_\limits{r=1}^n c \cdot h(r)$
$=a \sum_\limits{r=1}^n f(r)+b \sum_\limits{r=1}^n g(r)-c \sum_\limits{r=1}^n h(r)$
(iii) $\sum_\limits{r=1}^n r=\frac{n(n+1)}{2}, \sum_\limits{r=1}^n a=a r$ where is a constant.
2008
JEE Advanced
MSQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 2008 Paper 1 Offline
Let ${S_n} = \sum\limits_{k = 1}^n {{n \over {{n^2} + kn + {k^2}}}} $ and ${T_n} = \sum\limits_{k = 0}^{n - 1} {{n \over {{n^2} + kn + {k^2}}}} $ for $n$ $=1, 2, 3, ............$ Then,
A.
${S_n} < {\pi \over {3\sqrt 3 }}$
B.
${S_n} > {\pi \over {3\sqrt 3 }}$
C.
${T_n} < {\pi \over {3\sqrt 3 }}$
D.
${T_n} > {\pi \over {3\sqrt 3 }}$
Show Answer
Correct Answer: A,D
Explanation:
${S_n} < \mathop {\lim }\limits_{x \to \infty } {S_n} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{1 \over n}{1 \over {1 + {k \over n} + \left( {{k \over {{n^2}}}} \right)}}} $
$ = \int\limits_0^1 {{{dx} \over {1 + x + {x^2}}} = {\pi \over {3\sqrt 3 }}} $
As
$h\sum\limits_{k = 0}^n {f(kh) > \int\limits_0^1 {f(x)dx > h} } $
$\sum\limits_{k = 1}^n {f(kh)} $
So, ${T_n} > {\pi \over {3\sqrt 3 }}$
1999
JEE Advanced
MSQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1999
For a positive integer $n$, let
A.
$a\left( {100} \right) \le 100$
B.
$a\left( {100} \right) > 100$
C.
$a\left( {200} \right) \le 100$
D.
$a\left( {200} \right) > 100$
Show Answer
1993
JEE Advanced
MSQ
iCON Education HYD, 79930 92826, 73309 72826
IIT-JEE 1993
For $0 < \phi < \pi /2,$ if
Show Answer