Sequences and Series
293 Questions
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 27th June Morning Shift
If the sum of the first ten terms of the series
${1 \over 5} + {2 \over {65}} + {3 \over {325}} + {4 \over {1025}} + {5 \over {2501}} + \,\,....$
is ${m \over n}$, where m and n are co-prime numbers, then m + n is equal to ______________.
Correct Answer: 276
Explanation:
${T_r} = {r \over {{{(2{r^2})}^2} + 1}}$
$ = {r \over {{{(2{r^2} + 1)}^2} - {{(2r)}^2}}}$
$ = {1 \over 4}{{4r} \over {(2{r^2} + 2r + 1)(2{r^2} - 2r + 1)}}$
${S_{10}} = {1 \over 4}\sum\limits_{r = 1}^{10} {\left( {{1 \over {(2{r^2} - 2r + 1)}} - {1 \over {(2{r^2} + 2r + 1)}}} \right)} $
$ = {1 \over 4}\left[ {1 - {1 \over 5} + {1 \over 5} - {1 \over {13}} + \,\,....\,\, + \,\,{1 \over {181}} - {1 \over {221}}} \right]$
$ \Rightarrow {S_{10}} = {1 \over 4}\,.\,{{220} \over {221}} = {{55} \over {221}} = {m \over n}$
$\therefore$ $m + n = 276$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th June Evening Shift
If a1 (> 0), a2, a3, a4, a5 are in a G.P., a2 + a4 = 2a3 + 1 and 3a2 + a3 = 2a4, then a2 + a4 + 2a5 is equal to ___________.
Correct Answer: 40
Explanation:
Let G.P. be a1 = a, a2 = ar, a3 = ar2, .........
$\because$ 3a2 + a3 = 2a4
$\Rightarrow$ 3ar + ar2 = 2ar3
$\Rightarrow$ 2ar2 $-$ r $-$ 3 = 0
$\therefore$ r = $-$1 or ${3 \over 2}$
$\because$ a1 = a > 0 then r $\ne$ $-$1
Now, a2 + a4 = 2a3 + 1
ar + ar3 = 2ar2 + 1
$a\left( {{3 \over 2} + {{27} \over 8} - {9 \over 2}} \right) = 1$
$\therefore$ a = ${8 \over 3}$
$\therefore$ a2 + a4 + 2a5 = a(r + r3 + 2r4)
$ = {8 \over 3}\left( {{3 \over 2} + {{27} \over 8} + {{81} \over 8}} \right) = 40$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th June Morning Shift
For a natural number n, let ${\alpha _n} = {19^n} - {12^n}$. Then, the value of ${{31{\alpha _9} - {\alpha _{10}}} \over {57{\alpha _8}}}$ is ___________.
Correct Answer: 4
Explanation:
${\alpha _n} = {19^n} - {12^n}$
Let equation of roots 12 & 19 i.e.
${x^2} - 31x + 228 = 0$
$ \Rightarrow (31 - x) = {{228} \over x}$ (where x can be 19 or 12)
$\therefore$ ${{31{\alpha _9} - {\alpha _{10}}} \over {57{\alpha _8}}} = {{31({{19}^9} - {{12}^9}) - ({{19}^{10}} - {{12}^{10}})} \over {57({{19}^8} - {{12}^8})}}$
$ = {{{{19}^9}(31 - 19) - {{12}^9}(31 - 12)} \over {57({{19}^8} - {{12}^8})}}$
$ = {{228({{19}^8} - {{12}^8})} \over {57({{19}^8} - {{12}^8})}} = 4$.
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th June Morning Shift
The greatest integer less than or equal to the sum of first 100 terms of the sequence ${1 \over 3},{5 \over 9},{{19} \over {27}},{{65} \over {81}},$ ...... is equal to ___________.
Correct Answer: 98
Explanation:
$S = {1 \over 3} + {5 \over 9} + {{19} \over {27}} + {{65} \over {81}}\, + $ ....
$ = \sum\limits_{r = 1}^{100} {\left( {{{{3^r} - {2^r}} \over {{3^r}}}} \right)} $
$ = 100 - {2 \over 3}{{\left( {1 - {{\left( {{2 \over 3}} \right)}^{100}}} \right)} \over {1/3}}$
$ = 98 + 2{\left( {{2 \over 3}} \right)^{100}}$
$\therefore$ $[S] = 98$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 1st September Evening Shift
Let Sn = 1 . (n $-$ 1) + 2 . (n $-$ 2) + 3 . (n $-$ 3) + ..... + (n $-$ 1) . 1, n $\ge$ 4.
The sum $\sum\limits_{n = 4}^\infty {\left( {{{2{S_n}} \over {n!}} - {1 \over {(n - 2)!}}} \right)} $ is equal to :
The sum $\sum\limits_{n = 4}^\infty {\left( {{{2{S_n}} \over {n!}} - {1 \over {(n - 2)!}}} \right)} $ is equal to :
A.
${{e - 1} \over 3}$
B.
${{e - 2} \over 6}$
C.
${e \over 3}$
D.
${e \over 6}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 1st September Evening Shift
Let a1, a2, ..........., a21 be an AP such that $\sum\limits_{n = 1}^{20} {{1 \over {{a_n}{a_{n + 1}}}} = {4 \over 9}} $. If the sum of this AP is 189, then a6a16 is equal to :
A.
57
B.
72
C.
48
D.
36
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Evening Shift
Let a1, a2, a3, ..... be an A.P. If ${{{a_1} + {a_2} + .... + {a_{10}}} \over {{a_1} + {a_2} + .... + {a_p}}} = {{100} \over {{p^2}}}$, p $\ne$ 10, then ${{{a_{11}}} \over {{a_{10}}}}$ is equal to :
A.
${{19} \over {21}}$
B.
${{100} \over {121}}$
C.
${{21} \over {19}}$
D.
${{121} \over {100}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Morning Shift
The sum of 10 terms of the series
${3 \over {{1^2} \times {2^2}}} + {5 \over {{2^2} \times {3^2}}} + {7 \over {{3^2} \times {4^2}}} + ....$ is :
${3 \over {{1^2} \times {2^2}}} + {5 \over {{2^2} \times {3^2}}} + {7 \over {{3^2} \times {4^2}}} + ....$ is :
A.
1
B.
${{120} \over {121}}$
C.
${{99} \over {100}}$
D.
${{143} \over {144}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Morning Shift
Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d. If the fourth term of GP is 3 r2, then r2 $-$ d is equal to :
A.
7 $-$ 7$\sqrt 3 $
B.
7 + $\sqrt 3 $
C.
7 $-$ $\sqrt 3 $
D.
7 + 3$\sqrt 3 $
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th August Evening Shift
If 0 < x < 1 and $y = {1 \over 2}{x^2} + {2 \over 3}{x^3} + {3 \over 4}{x^4} + ....$, then the value of e1 + y at $x = {1 \over 2}$ is :
A.
${1 \over 2}{e^2}$
B.
2e
C.
${1 \over 2}\sqrt e $
D.
2e2
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th August Morning Shift
If 0 < x < 1, then ${3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + .....$, is equal to :
A.
$x\left( {{{1 + x} \over {1 - x}}} \right) + {\log _e}(1 - x)$
B.
$x\left( {{{1 - x} \over {1 + x}}} \right) + {\log _e}(1 - x)$
C.
${{1 - x} \over {1 + x}} + {\log _e}(1 - x)$
D.
${{1 + x} \over {1 - x}} + {\log _e}(1 - x)$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th August Morning Shift
If for x, y $\in$ R, x > 0, y = log10x + log10x1/3 + log10x1/9 + ...... upto $\infty$ terms
and ${{2 + 4 + 6 + .... + 2y} \over {3 + 6 + 9 + ..... + 3y}} = {4 \over {{{\log }_{10}}x}}$, then the ordered pair (x, y) is equal to :
and ${{2 + 4 + 6 + .... + 2y} \over {3 + 6 + 9 + ..... + 3y}} = {4 \over {{{\log }_{10}}x}}$, then the ordered pair (x, y) is equal to :
A.
(106, 6)
B.
(104, 6)
C.
(102, 3)
D.
(106, 9)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Morning Shift
The sum of the series
${1 \over {x + 1}} + {2 \over {{x^2} + 1}} + {{{2^2}} \over {{x^4} + 1}} + ...... + {{{2^{100}}} \over {{x^{{2^{100}}}} + 1}}$ when x = 2 is :
${1 \over {x + 1}} + {2 \over {{x^2} + 1}} + {{{2^2}} \over {{x^4} + 1}} + ...... + {{{2^{100}}} \over {{x^{{2^{100}}}} + 1}}$ when x = 2 is :
A.
$1 + {{{2^{101}}} \over {{4^{101}} - 1}}$
B.
$1 + {{{2^{100}}} \over {{4^{101}} - 1}}$
C.
$1 - {{{2^{100}}} \over {{4^{100}} - 1}}$
D.
$1 - {{{2^{101}}} \over {{2^{400}} - 1}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Morning Shift
If the sum of an infinite GP a, ar, ar2, ar3, ....... is 15 and the sum of the squares of its each term is 150, then the sum of ar2, ar4, ar6, ....... is :
A.
${5 \over 2}$
B.
${1 \over 2}$
C.
${25 \over 2}$
D.
${9 \over 2}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Morning Shift
Let Sn be the sum of the first n terms of an arithmetic progression. If S3n = 3S2n, then the value of ${{{S_{4n}}} \over {{S_{2n}}}}$ is :
A.
6
B.
4
C.
2
D.
8
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 22th July Evening Shift
Let Sn denote the sum of first n-terms of an arithmetic progression. If S10 = 530, S5 = 140, then S20 $-$ S6 is equal to:
A.
1862
B.
1842
C.
1852
D.
1872
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Evening Shift
If sum of the first 21 terms of the series ${\log _{{9^{1/2}}}}x + {\log _{{9^{1/3}}}}x + {\log _{{9^{1/4}}}}x + .......$, where x > 0 is 504, then x is equal to
A.
243
B.
9
C.
7
D.
81
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Evening Shift
Let S1 be the sum of first 2n terms of an arithmetic progression. Let S2 be the sum of first 4n terms of the same arithmetic progression. If (S2 $-$ S1) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to :
A.
7000
B.
1000
C.
3000
D.
5000
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
If $\alpha$, $\beta$ are natural numbers such that
100$\alpha$ $-$ 199$\beta$ = (100)(100) + (99)(101) + (98)(102) + ...... + (1)(199), then the slope of the line passing through ($\alpha$, $\beta$) and origin is :
100$\alpha$ $-$ 199$\beta$ = (100)(100) + (99)(101) + (98)(102) + ...... + (1)(199), then the slope of the line passing through ($\alpha$, $\beta$) and origin is :
A.
540
B.
550
C.
530
D.
510
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
${1 \over {{3^2} - 1}} + {1 \over {{5^2} - 1}} + {1 \over {{7^2} - 1}} + .... + {1 \over {{{(201)}^2} - 1}}$ is equal to
A.
${{101} \over {404}}$
B.
${{25} \over {101}}$
C.
${{101} \over {408}}$
D.
${{99} \over {400}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Evening Shift
The sum of the series
$\sum\limits_{n = 1}^\infty {{{{n^2} + 6n + 10} \over {(2n + 1)!}}} $ is equal to :
$\sum\limits_{n = 1}^\infty {{{{n^2} + 6n + 10} \over {(2n + 1)!}}} $ is equal to :
A.
${{41} \over 8}e + {{19} \over 8}{e^{ - 1}} - 10$
B.
${{41} \over 8}e - {{19} \over 8}{e^{ - 1}} - 10$
C.
${{41} \over 8}e + {{19} \over 8}{e^{ - 1}} + 10$
D.
$ - {{41} \over 8}e + {{19} \over 8}{e^{ - 1}} - 10$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
The sum of the infinite series
$1 + {2 \over 3} + {7 \over {{3^2}}} + {{12} \over {{3^3}}} + {{17} \over {{3^4}}} + {{22} \over {{3^5}}} + ......$ is equal to :
$1 + {2 \over 3} + {7 \over {{3^2}}} + {{12} \over {{3^3}}} + {{17} \over {{3^4}}} + {{22} \over {{3^5}}} + ......$ is equal to :
A.
${9 \over 4}$
B.
${13 \over 4}$
C.
${15 \over 4}$
D.
${11 \over 4}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
In an increasing geometric series, the sum of the second and the sixth term is ${{25} \over 2}$ and the product of the third and fifth term is 25. Then, the sum of 4th, 6th and 8th terms is equal to :
A.
30
B.
32
C.
26
D.
35
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
The minimum value of $f(x) = {a^{{a^x}}} + {a^{1 - {a^x}}}$, where a, $x \in R$ and a > 0, is equal to :
A.
$a + {1 \over a}$
B.
2a
C.
a + 1
D.
$2\sqrt a $
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
If $0 < \theta ,\phi < {\pi \over 2},x = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } ,y = \sum\limits_{n = 0}^\infty {{{\sin }^{2n}}\phi } $ and $z = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta .{{\sin }^{2n}}\phi } $ then :
A.
xy $-$ z = (x + y)z
B.
xyz = 4
C.
xy + z = (x + y)z
D.
xy + yz + zx = z
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Evening Shift
The number of 4-digit numbers which are neither multiple of 7 nor multiple of 3 is ____________.
Correct Answer: 5143
Explanation:
A = 4-digit numbers divisible by 3
A = 1002, 1005, ....., 9999.
9999 = 1002 + (n $-$ 1)3
$\Rightarrow$ (n $-$ 1)3 = 8997 $\Rightarrow$ n = 3000
B = 4-digit numbers divisible by 7
B = 1001, 1008, ......., 9996
$\Rightarrow$ 9996 = 1001 + (n $-$ 1)7
$\Rightarrow$ n = 1286
A $\cap$ B = 1008, 1029, ....., 9996
9996 = 1008 + (n $-$ 1)21
$\Rightarrow$ n = 429
So, no divisible by either 3 or 7
= 3000 + 1286 $-$ 429 = 3857
total 4-digits numbers = 9000
required numbers = 9000 $-$ 3857 = 5143
A = 1002, 1005, ....., 9999.
9999 = 1002 + (n $-$ 1)3
$\Rightarrow$ (n $-$ 1)3 = 8997 $\Rightarrow$ n = 3000
B = 4-digit numbers divisible by 7
B = 1001, 1008, ......., 9996
$\Rightarrow$ 9996 = 1001 + (n $-$ 1)7
$\Rightarrow$ n = 1286
A $\cap$ B = 1008, 1029, ....., 9996
9996 = 1008 + (n $-$ 1)21
$\Rightarrow$ n = 429
So, no divisible by either 3 or 7
= 3000 + 1286 $-$ 429 = 3857
total 4-digits numbers = 9000
required numbers = 9000 $-$ 3857 = 5143
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Evening Shift
If $S = {7 \over 5} + {9 \over {{5^2}}} + {{13} \over {{5^3}}} + {{19} \over {{5^4}}} + ....$, then 160 S is equal to ________.
Correct Answer: 305
Explanation:
$S = {7 \over 5} + {9 \over {{5^2}}} + {{13} \over {{5^3}}} + {{19} \over {{5^4}}} + ....$
${1 \over 5}S = {7 \over 5} + {9 \over {{5^3}}} + {{13} \over {{5^4}}} + ....$
On subtracting
${4 \over 5}S = {7 \over 5} + {2 \over {{5^2}}} + {4 \over {{5^3}}} + {6 \over {{5^4}}} + ....$
$S = {7 \over {14}} + {1 \over {10}}\left( {1 + {2 \over 5} + {3 \over {{5^2}}} + ...} \right)$
$S = {7 \over 4} + {1 \over {10}}{\left( {1 - {1 \over 5}} \right)^{ - 2}}$
$ = {7 \over 4} + {1 \over {10}} \times {{25} \over {16}} = {{61} \over {32}}$
$\Rightarrow$ 160S = 5 $\times$ 61 = 305
${1 \over 5}S = {7 \over 5} + {9 \over {{5^3}}} + {{13} \over {{5^4}}} + ....$
On subtracting
${4 \over 5}S = {7 \over 5} + {2 \over {{5^2}}} + {4 \over {{5^3}}} + {6 \over {{5^4}}} + ....$
$S = {7 \over {14}} + {1 \over {10}}\left( {1 + {2 \over 5} + {3 \over {{5^2}}} + ...} \right)$
$S = {7 \over 4} + {1 \over {10}}{\left( {1 - {1 \over 5}} \right)^{ - 2}}$
$ = {7 \over 4} + {1 \over {10}} \times {{25} \over {16}} = {{61} \over {32}}$
$\Rightarrow$ 160S = 5 $\times$ 61 = 305
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Evening Shift
The sum of all 3-digit numbers less than or equal to 500, that are formed without using the digit "1" and they all are multiple of 11, is _____________.
Correct Answer: 7744
Explanation:
209, 220, 231, ..........., 495
Sum = ${{27} \over 2}$(209 + 495) = 9504
Number containing 1 at unit place $\matrix{ {\underline 2 } & {\underline 3 } & {\underline 1 } \cr {\underline 3 } & {\underline 4 } & {\underline 1 } \cr {\underline 4 } & {\underline 5 } & {\underline 1 } \cr } $
Number containing 1 at 10th place $\matrix{ {\underline 3 } & {\underline 1 } & {\underline 9 } \cr {\underline 4 } & {\underline 1 } & {\underline 8 } \cr } $
Required = 9504 $-$ (231 + 341 + 451 + 319 + 418)
= 7744
Sum = ${{27} \over 2}$(209 + 495) = 9504
Number containing 1 at unit place $\matrix{ {\underline 2 } & {\underline 3 } & {\underline 1 } \cr {\underline 3 } & {\underline 4 } & {\underline 1 } \cr {\underline 4 } & {\underline 5 } & {\underline 1 } \cr } $
Number containing 1 at 10th place $\matrix{ {\underline 3 } & {\underline 1 } & {\underline 9 } \cr {\underline 4 } & {\underline 1 } & {\underline 8 } \cr } $
Required = 9504 $-$ (231 + 341 + 451 + 319 + 418)
= 7744
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Evening Shift
Let a1, a2, ......., a10 be an AP with common difference $-$ 3 and b1, b2, ........., b10 be a GP with common ratio 2. Let ck = ak + bk, k = 1, 2, ......, 10. If c2 = 12 and c3 = 13, then $\sum\limits_{k = 1}^{10} {{c_k}} $ is equal to _________.
Correct Answer: 2021
Explanation:
$a_{1}, a_{2}, a_{3}, \ldots, a_{10}$ are in AP common difference $=-3$
$b_{1}, b_{2}, b_{3}, \ldots, b_{10}$ are in GP common ratio $=2$
Since, $c_{k}=a_{k}+b_{k}, k=1,2,3 \ldots \ldots, 10$
$\therefore c_{2} =a_{2}+b_{2}=12$
$ c_{3} =a_{3}+b_{3}=13$
Now, $\mathrm{C}_{3}-\mathrm{C}_{2}=1$
$ \begin{array}{ll} \Rightarrow & \left(a_{3}-a_{2}\right)+\left(b_{3}-b_{2}\right) \neq 1 \Rightarrow-3+\left(2 b_{2}-b_{2}\right) \neq 1 \\ \Rightarrow & b_{2}=4 \\ \therefore & a_{2}=8 \end{array} $
So, AP is $11,8,5, \ldots$.
Now, $\sum_{k=1}^{10} C_{k}=\sum_{k=1}^{10} a_{k}+\sum_{k=1}^{10} b_{k}$
$ \begin{aligned} &=\left(\frac{10}{2}\right)[22+9(-3)]+2\left(\frac{2^{10}-1}{2-1}\right) \\ &=5(22-27)+2(1023)=2046-25 \\ &=2021 \end{aligned} $
$b_{1}, b_{2}, b_{3}, \ldots, b_{10}$ are in GP common ratio $=2$
Since, $c_{k}=a_{k}+b_{k}, k=1,2,3 \ldots \ldots, 10$
$\therefore c_{2} =a_{2}+b_{2}=12$
$ c_{3} =a_{3}+b_{3}=13$
Now, $\mathrm{C}_{3}-\mathrm{C}_{2}=1$
$ \begin{array}{ll} \Rightarrow & \left(a_{3}-a_{2}\right)+\left(b_{3}-b_{2}\right) \neq 1 \Rightarrow-3+\left(2 b_{2}-b_{2}\right) \neq 1 \\ \Rightarrow & b_{2}=4 \\ \therefore & a_{2}=8 \end{array} $
So, AP is $11,8,5, \ldots$.
Now, $\sum_{k=1}^{10} C_{k}=\sum_{k=1}^{10} a_{k}+\sum_{k=1}^{10} b_{k}$
$ \begin{aligned} &=\left(\frac{10}{2}\right)[22+9(-3)]+2\left(\frac{2^{10}-1}{2-1}\right) \\ &=5(22-27)+2(1023)=2046-25 \\ &=2021 \end{aligned} $
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Morning Shift
If ${\log _3}2,{\log _3}({2^x} - 5),{\log _3}\left( {{2^x} - {7 \over 2}} \right)$ are in an arithmetic progression, then the value of x is equal to _____________.
Correct Answer: 3
Explanation:
$2{\log _3}({2^x} - 5) = {\log _2} + {\log _3}\left( {{2^x} - {7 \over 2}} \right)$
Let ${2^x} = t$
${\log _3}{(t - 5)^2} = {\log _3}2\left( {t - {7 \over 2}} \right)$
${(t - 5)^2} = 2t - 7$
${t^2} - 12t + 32 = 0$
$(t - 4)(t - 8) = 0$
$\Rightarrow$ 2x = 4 or 2x = 8
x = 2 (Rejected)
Or x = 3
Let ${2^x} = t$
${\log _3}{(t - 5)^2} = {\log _3}2\left( {t - {7 \over 2}} \right)$
${(t - 5)^2} = 2t - 7$
${t^2} - 12t + 32 = 0$
$(t - 4)(t - 8) = 0$
$\Rightarrow$ 2x = 4 or 2x = 8
x = 2 (Rejected)
Or x = 3
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Morning Shift
If the value of
${\left( {1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + ....upto\,\infty } \right)^{{{\log }_{(0.25)}}\left( {{1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....upto\,\infty } \right)}}$
is $l$, then $l$2 is equal to _______________.
${\left( {1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + ....upto\,\infty } \right)^{{{\log }_{(0.25)}}\left( {{1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....upto\,\infty } \right)}}$
is $l$, then $l$2 is equal to _______________.
Correct Answer: 3
Explanation:
$l = {\left( {\underbrace {1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}}}_S + ....} \right)^{{{\log }_{0.25}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ...} \right)}}$
$S = 1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + ....$
${S \over 3} = {1 \over 3} + {2 \over {{3^2}}} + {6 \over {{3^3}}} + .....$${{2x} \over 3} = 1 + {1 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + ....$
${{2S} \over 3} = {4 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + .....$
$S = {3 \over 2}\left( {{{4/3} \over {1 - 1/3}}} \right) = 3$
Now, $l = {\left( 3 \right)^{{{\log }_{0.25}}\left( {{{1/3} \over {1 - 1/3}}} \right)}}$
$l = {3^{{{\log }_{\left( {(1/4)} \right)}}\left( {{1 \over 2}} \right)}} = {3^{1/2}} = \sqrt 3 $
$\Rightarrow$ l2 = 3
$S = 1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + ....$
${S \over 3} = {1 \over 3} + {2 \over {{3^2}}} + {6 \over {{3^3}}} + .....$${{2x} \over 3} = 1 + {1 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + ....$
${{2S} \over 3} = {4 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + .....$
$S = {3 \over 2}\left( {{{4/3} \over {1 - 1/3}}} \right) = 3$
Now, $l = {\left( 3 \right)^{{{\log }_{0.25}}\left( {{{1/3} \over {1 - 1/3}}} \right)}}$
$l = {3^{{{\log }_{\left( {(1/4)} \right)}}\left( {{1 \over 2}} \right)}} = {3^{1/2}} = \sqrt 3 $
$\Rightarrow$ l2 = 3
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 22th July Evening Shift
The sum of all the elements in the set {n$\in$ {1, 2, ....., 100} | H.C.F. of n and 2040 is 1} is equal to _____________.
Correct Answer: 1251
Explanation:
2040 = 23 $\times$ 3 $\times$ 5 $\times$ 17
n should not be multiple of 2, 3, 5 and 17.
Sum of all n = (1 + 3 + 5 + ...... + 99) $-$ (3 + 9 + 15 + 21 + ...... + 99) $-$ (5 + 25 + 35 + 55 + 65 + 85 + 95) $-$ (17)
= 2500 $-$ ${{17} \over 2}$(3 + 99) $-$ 365 $-$ 17
2500 $-$ 867 $-$ 365 $-$ 17
= 1251
n should not be multiple of 2, 3, 5 and 17.
Sum of all n = (1 + 3 + 5 + ...... + 99) $-$ (3 + 9 + 15 + 21 + ...... + 99) $-$ (5 + 25 + 35 + 55 + 65 + 85 + 95) $-$ (17)
= 2500 $-$ ${{17} \over 2}$(3 + 99) $-$ 365 $-$ 17
2500 $-$ 867 $-$ 365 $-$ 17
= 1251
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Evening Shift
For k $\in$ N, let ${1 \over {\alpha (\alpha + 1)(\alpha + 2).........(\alpha + 20)}} = \sum\limits_{K = 0}^{20} {{{{A_k}} \over {\alpha + k}}} $, where $\alpha > 0$. Then the value of $100{\left( {{{{A_{14}} + {A_{15}}} \over {{A_{13}}}}} \right)^2}$ is equal to _____________.
Correct Answer: 9
Explanation:
${1 \over {\alpha (\alpha + 1)(\alpha + 2).........(\alpha + 20)}} = \sum\limits_{K = 0}^{20} {{{{A_k}} \over {\alpha + k}}} $
${A_{14}} = {1 \over {( - 14)( - 13)......( - 1)(1).......(6)}} = {1 \over {14!.6!}}$
${A_{15}} = {1 \over {( - 15)( - 14)......( - 1)(1).......(5)}} = {1 \over {15!.5!}}$
${A_{13}} = {1 \over {( - 13)......( - 1)(1).......(7)}} = {1 \over {13!.7!}}$
${{{A_{14}}} \over {{A_{13}}}} = {1 \over {14!.6!}} \times - 13! \times 7! = {{ - 7} \over {14}} = - {1 \over 2}$
${{{A_{15}}} \over {{A_{13}}}} = {1 \over {15! \times 5!}} \times - 13! \times 7! = {{42} \over {15 \times 14}} = {1 \over 5}$
$100{\left( {{{{A_{14}}} \over {{A_{13}}}} + {{{A_{15}}} \over {{A_{13}}}}} \right)^2} = 100{\left( { - {1 \over 2} + {1 \over 5}} \right)^2} = 9$
${A_{14}} = {1 \over {( - 14)( - 13)......( - 1)(1).......(6)}} = {1 \over {14!.6!}}$
${A_{15}} = {1 \over {( - 15)( - 14)......( - 1)(1).......(5)}} = {1 \over {15!.5!}}$
${A_{13}} = {1 \over {( - 13)......( - 1)(1).......(7)}} = {1 \over {13!.7!}}$
${{{A_{14}}} \over {{A_{13}}}} = {1 \over {14!.6!}} \times - 13! \times 7! = {{ - 7} \over {14}} = - {1 \over 2}$
${{{A_{15}}} \over {{A_{13}}}} = {1 \over {15! \times 5!}} \times - 13! \times 7! = {{42} \over {15 \times 14}} = {1 \over 5}$
$100{\left( {{{{A_{14}}} \over {{A_{13}}}} + {{{A_{15}}} \over {{A_{13}}}}} \right)^2} = 100{\left( { - {1 \over 2} + {1 \over 5}} \right)^2} = 9$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Evening Shift
Let $\left\{ {{a_n}} \right\}_{n = 1}^\infty $ be a sequence such that a1 = 1, a2 = 1 and ${a_{n + 2}} = 2{a_{n + 1}} + {a_n}$ for all n $\ge$ 1. Then the value of $47\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{2^{3n}}}}} $ is equal to ______________.
Correct Answer: 7
Explanation:
${a_{n + 2}} = 2{a_{n + 1}} + {a_n}$, let $\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{8^n}}}} = P$
Divide by 8n we get
${{{a_{n + 2}}} \over {{8^n}}} = {{2{a_{n + 1}}} \over {{8^n}}} + {{{a_n}} \over {{8^n}}}$
$ \Rightarrow 64{{{a_{n + 2}}} \over {{8^{n + 2}}}} = {{16{a_{n + 1}}} \over {{8^{n + 1}}}} + {{{a_n}} \over {{8^n}}}$
$64\sum\limits_{n = 1}^\infty {{{{a_{n + 2}}} \over {{8^{n + 2}}}}} = 16\sum\limits_{n = 1}^\infty {{{{a_{n + 1}}} \over {{8^{n + 1}}}}} + \sum\limits_{n = 1}^\infty {{{{a_n}} \over {{8^n}}}} $
$64\left( {P - {{{a_1}} \over 8} - {{{a_2}} \over {{8^2}}}} \right) = 16\left( {P - {{{a_1}} \over 8}} \right) + P$
$ \Rightarrow 64\left( {P - {1 \over 8} - {1 \over {64}}} \right) = 16\left( {P - {1 \over 8}} \right) + P$
$64P - 8 - 1 = 16P - 2 + P$
$47P = 7$
Divide by 8n we get
${{{a_{n + 2}}} \over {{8^n}}} = {{2{a_{n + 1}}} \over {{8^n}}} + {{{a_n}} \over {{8^n}}}$
$ \Rightarrow 64{{{a_{n + 2}}} \over {{8^{n + 2}}}} = {{16{a_{n + 1}}} \over {{8^{n + 1}}}} + {{{a_n}} \over {{8^n}}}$
$64\sum\limits_{n = 1}^\infty {{{{a_{n + 2}}} \over {{8^{n + 2}}}}} = 16\sum\limits_{n = 1}^\infty {{{{a_{n + 1}}} \over {{8^{n + 1}}}}} + \sum\limits_{n = 1}^\infty {{{{a_n}} \over {{8^n}}}} $
$64\left( {P - {{{a_1}} \over 8} - {{{a_2}} \over {{8^2}}}} \right) = 16\left( {P - {{{a_1}} \over 8}} \right) + P$
$ \Rightarrow 64\left( {P - {1 \over 8} - {1 \over {64}}} \right) = 16\left( {P - {1 \over 8}} \right) + P$
$64P - 8 - 1 = 16P - 2 + P$
$47P = 7$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
Sn(x) = loga1/2x + loga1/3x + loga1/6x + loga1/11x + loga1/18x + loga1/27x + ...... up to n-terms, where a > 1. If S24(x) = 1093 and S12(2x) = 265, then value of a is equal to ____________.
Correct Answer: 16
Explanation:
${S_n}(x) = {\log _a}{x^2} + {\log _a}{x^3} + {\log _a}{x^6} + {\log _a}{x^{11}}$
${S_n}(x) = 2{\log _a}x + 3{\log _a}x + 6{\log _a}x + 11{\log _a}x + ......$
${S_n}(x) = {\log _a}x(2 + 3 + 6 + 11 + .....)$
${S_r} = 2 + 3 + 6 + 11$
$ \therefore $ Tn = 2 + (1 + 3 + 5 +......+ (n - 1))
= 2 + ${{n - 1} \over 2}\left[ {2.1 + \left( {n - 2} \right)2} \right]$
= 2 + $\left( {n - 1} \right)\left[ {1 + \left( {n - 2} \right)} \right]$
= n2 - 2n + 3
General term ${T_r} = {r^2} - 2r + 3$
${S_n}(x) = \sum\limits_{r = 1}^n {{{\log }_a}x({r^2} - 2r + 3)} $
${S_{24}}(x) = \sum\limits_{r = 1}^{24} {{{\log }_a}x({r^2} - 2r + 3)} $
${S_{24}}(x) = {\log _a}x\sum\limits_{r = 1}^{24} {({r^2} - 2r + 3)} $
$1093 = 4372{\log _a}x$
${\log _a}x = {1 \over 4}$
$x = {a^{1/4}}$ .....(i)
${S_{12}}(2x) = {\log _a}(2x)\sum\limits_{r = 1}^{12} {({r^2} - 2r + 3)} $
$265 = 530{\log _a}(2x)$
${\log _a}(2x) = {1 \over 2}$
$2x = {a^{1/2}}$ ....(ii)
From (i) and (ii), we get
$2{a^{{1 \over 4}}} = {a^{{1 \over 2}}}$
$ \Rightarrow $ ${\left( {2{a^{{1 \over 4}}}} \right)^4} = {\left( {{a^{{1 \over 2}}}} \right)^4}$
$ \Rightarrow $ 16$a$ = $a$2
$ \Rightarrow $ $a = 16$
${S_n}(x) = 2{\log _a}x + 3{\log _a}x + 6{\log _a}x + 11{\log _a}x + ......$
${S_n}(x) = {\log _a}x(2 + 3 + 6 + 11 + .....)$
${S_r} = 2 + 3 + 6 + 11$
$ \therefore $ Tn = 2 + (1 + 3 + 5 +......+ (n - 1))
= 2 + ${{n - 1} \over 2}\left[ {2.1 + \left( {n - 2} \right)2} \right]$
= 2 + $\left( {n - 1} \right)\left[ {1 + \left( {n - 2} \right)} \right]$
= n2 - 2n + 3
General term ${T_r} = {r^2} - 2r + 3$
${S_n}(x) = \sum\limits_{r = 1}^n {{{\log }_a}x({r^2} - 2r + 3)} $
${S_{24}}(x) = \sum\limits_{r = 1}^{24} {{{\log }_a}x({r^2} - 2r + 3)} $
${S_{24}}(x) = {\log _a}x\sum\limits_{r = 1}^{24} {({r^2} - 2r + 3)} $
$1093 = 4372{\log _a}x$
${\log _a}x = {1 \over 4}$
$x = {a^{1/4}}$ .....(i)
${S_{12}}(2x) = {\log _a}(2x)\sum\limits_{r = 1}^{12} {({r^2} - 2r + 3)} $
$265 = 530{\log _a}(2x)$
${\log _a}(2x) = {1 \over 2}$
$2x = {a^{1/2}}$ ....(ii)
From (i) and (ii), we get
$2{a^{{1 \over 4}}} = {a^{{1 \over 2}}}$
$ \Rightarrow $ ${\left( {2{a^{{1 \over 4}}}} \right)^4} = {\left( {{a^{{1 \over 2}}}} \right)^4}$
$ \Rightarrow $ 16$a$ = $a$2
$ \Rightarrow $ $a = 16$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
Let ${1 \over {16}}$, a and b be in G.P. and ${1 \over a}$, ${1 \over b}$, 6 be in A.P., where a, b > 0. Then 72(a + b) is equal to ___________.
Correct Answer: 14
Explanation:
${a^2} = {b \over {16}}$ and ${2 \over b} = {1 \over a} + 6$
Solving, we get $a = {1 \over {12}}$ or $a = - {1 \over 4}$ [rejected]
if $a = {1 \over {12}} \Rightarrow b = {1 \over 9}$
$ \therefore $ $72(a + b) = 72\left( {{1 \over {12}} + {1 \over 9}} \right) = 14$
Solving, we get $a = {1 \over {12}}$ or $a = - {1 \over 4}$ [rejected]
if $a = {1 \over {12}} \Rightarrow b = {1 \over 9}$
$ \therefore $ $72(a + b) = 72\left( {{1 \over {12}} + {1 \over 9}} \right) = 14$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
Consider an arithmetic series and a geometric series having four initial terms from the set {11, 8, 21, 16, 26, 32, 4}. If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these two series is equal to ___________.
Correct Answer: 3
Explanation:
A.P. from the set will be 11, 16, 21, 26 .....
G.P. from the set will be 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192 .....
So common terms are 16, 256, 4096.
G.P. from the set will be 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192 .....
So common terms are 16, 256, 4096.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
The total number of 4-digit numbers whose greatest common divisor with 18 is 3, is _________.
Correct Answer: 1000
Explanation:
Let N be the four digit number
gcd(N, 18) = 3
Hence N is an odd integer which is divisible by 3 but not by 9.
4 digit odd multiples of 3
1005, 1011, ..........., 9999 $ \to $ 1500
4 digit odd multiples of 9
1017, 1035, ..........., 9999 $ \to $ 500
Hence number of such N = 1000
gcd(N, 18) = 3
Hence N is an odd integer which is divisible by 3 but not by 9.
4 digit odd multiples of 3
1005, 1011, ..........., 9999 $ \to $ 1500
4 digit odd multiples of 9
1017, 1035, ..........., 9999 $ \to $ 500
Hence number of such N = 1000
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
If the arithmetic mean and geometric mean of the pth and qth terms of the
sequence $-$16, 8, $-$4, 2, ...... satisfy the equation
4x2 $-$ 9x + 5 = 0, then p + q is equal to __________.
sequence $-$16, 8, $-$4, 2, ...... satisfy the equation
4x2 $-$ 9x + 5 = 0, then p + q is equal to __________.
Correct Answer: 10
Explanation:
Given, $4{x^2} - 9x + 5 = 0$
$ \Rightarrow (x - 1)(4x - 5) = 0$
$ \Rightarrow $ A. M. $ = {5 \over 4}$, G. M. = 1 (As A. M. $ \ge $ G. M)
Again, for the series
$-$16, 8, $-$4, 2 ..........
${p^{th}}$ term ${t_p} = - 16{\left( {{{ - 1} \over 2}} \right)^{p - 1}}$
${q^{th}}$ term ${t_p} = 16{\left( {{{ - 1} \over 2}} \right)^{q - 1}}$
Now, A. M. = ${{{t_p} + {t_q}} \over 2} = {5 \over 4}$ & G. M. = $\sqrt {{t_p}{t_q}} = 1$
$ \Rightarrow {16^2}{\left( { - {1 \over 2}} \right)^{p + q - 2}} = 1$
$ \Rightarrow {( - 2)^8} = {( - 2)^{(p + q - 2)}}$
$ \Rightarrow p + q = 10$
$ \Rightarrow (x - 1)(4x - 5) = 0$
$ \Rightarrow $ A. M. $ = {5 \over 4}$, G. M. = 1 (As A. M. $ \ge $ G. M)
Again, for the series
$-$16, 8, $-$4, 2 ..........
${p^{th}}$ term ${t_p} = - 16{\left( {{{ - 1} \over 2}} \right)^{p - 1}}$
${q^{th}}$ term ${t_p} = 16{\left( {{{ - 1} \over 2}} \right)^{q - 1}}$
Now, A. M. = ${{{t_p} + {t_q}} \over 2} = {5 \over 4}$ & G. M. = $\sqrt {{t_p}{t_q}} = 1$
$ \Rightarrow {16^2}{\left( { - {1 \over 2}} \right)^{p + q - 2}} = 1$
$ \Rightarrow {( - 2)^8} = {( - 2)^{(p + q - 2)}}$
$ \Rightarrow p + q = 10$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Morning Shift
Let A1, A2, A3, ....... be squares such that for each n $ \ge $ 1, the length of the side of An equals the length of diagonal of An+1. If the length of A1 is 12 cm, then the smallest value of n for which area of An is less than one, is __________.
Correct Answer: 9
Explanation:
$ \therefore $ Side lengths are in G.P.
${T_n} = {{12} \over {{{\left( {\sqrt 2 } \right)}^{n - 1}}}}$
$ \therefore $ Area $ = {{144} \over {{2^{n - 1}} }}$ < 1
$ \Rightarrow {2^{n - 1}} > 144$
Smallest n = 9
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
The sum of first four terms of a geometric progression (G. P.) is ${{65} \over {12}}$ and the sum of their respective reciprocals is ${{65} \over {18}}$. If the product of first three terms of the G.P. is 1, and the third term is $\alpha$, then 2$\alpha$ is _________.
Correct Answer: 3
Explanation:
Let the terms are $a,ar,a{r^2},a{r^3}$
$a + ar + a{r^2} + a{r^3} = {{65} \over {12}}$ ..........(1)
${1 \over a} + {1 \over {ar}} + {1 \over {a{r^2}}} + {1 \over {a{r^3}}} = {{65} \over {18}}$
${1 \over a}\left( {{{{r^3} + {r^2} + r + 1} \over {{r^3}}}} \right) = {{65} \over {18}}$ ...............(2)
Doing ${{(1)} \over {(2)}},$
${a^2}{r^3} = {{18} \over {12}} = {3 \over 2}$
Also given, ${a^3}{r^3} = 1 \Rightarrow a\left( {{3 \over 2}} \right) = 1 \Rightarrow a = {2 \over 3}$
${4 \over 9}{r^3} = {3 \over 2} \Rightarrow {r^3} = {{{3^3}} \over {{2^3}}} \Rightarrow r = {3 \over 2}$
$\alpha = a{r^2} = {2 \over 3}.{\left( {{3 \over 2}} \right)^2} = {3 \over 2}$
$2\alpha = 3$
$a + ar + a{r^2} + a{r^3} = {{65} \over {12}}$ ..........(1)
${1 \over a} + {1 \over {ar}} + {1 \over {a{r^2}}} + {1 \over {a{r^3}}} = {{65} \over {18}}$
${1 \over a}\left( {{{{r^3} + {r^2} + r + 1} \over {{r^3}}}} \right) = {{65} \over {18}}$ ...............(2)
Doing ${{(1)} \over {(2)}},$
${a^2}{r^3} = {{18} \over {12}} = {3 \over 2}$
Also given, ${a^3}{r^3} = 1 \Rightarrow a\left( {{3 \over 2}} \right) = 1 \Rightarrow a = {2 \over 3}$
${4 \over 9}{r^3} = {3 \over 2} \Rightarrow {r^3} = {{{3^3}} \over {{2^3}}} \Rightarrow r = {3 \over 2}$
$\alpha = a{r^2} = {2 \over 3}.{\left( {{3 \over 2}} \right)^2} = {3 \over 2}$
$2\alpha = 3$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
The common difference of the A.P.
b1, b2, … , bm is 2 more than the common
difference of A.P. a1, a2, …, an. If
a40 = –159, a100 = –399 and b100 = a70, then b1 is equal to :
b1, b2, … , bm is 2 more than the common
difference of A.P. a1, a2, …, an. If
a40 = –159, a100 = –399 and b100 = a70, then b1 is equal to :
A.
127
B.
81
C.
–127
D.
-81
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
Let a , b, c , d and p be any non zero distinct real numbers such that
(a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0. Then :
(a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0. Then :
A.
a, c, p are in G.P.
B.
a, b, c, d are in G.P.
C.
a, b, c, d are in A.P.
D.
a, c, p are in A.P.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
If the sum of the first 20 terms of the series
${\log _{\left( {{7^{1/2}}} \right)}}x + {\log _{\left( {{7^{1/3}}} \right)}}x + {\log _{\left( {{7^{1/4}}} \right)}}x + ...$ is 460,
then x is equal to :
${\log _{\left( {{7^{1/2}}} \right)}}x + {\log _{\left( {{7^{1/3}}} \right)}}x + {\log _{\left( {{7^{1/4}}} \right)}}x + ...$ is 460,
then x is equal to :
A.
e2
B.
71/2
C.
72
D.
746/21
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
If the sum of the second, third and fourth terms
of a positive term G.P. is 3 and the sum of its
sixth, seventh and eighth terms is 243, then the
sum of the first 50 terms of this G.P. is :
A.
${2 \over {13}}\left( {{3^{50}} - 1} \right)$
B.
${1 \over {13}}\left( {{3^{50}} - 1} \right)$
C.
${1 \over {26}}\left( {{3^{49}} - 1} \right)$
D.
${1 \over {26}}\left( {{3^{50}} - 1} \right)$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
If ${3^{2\sin 2\alpha - 1}}$, 14 and ${3^{4 - 2\sin 2\alpha }}$ are the first three terms of an A.P. for some $\alpha $, then the sixth
terms of this A.P. is:
A.
66
B.
81
C.
65
D.
78
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
If 210 + 29.31 + 28
.32 +.....+ 2.39 + 310 = S - 211, then S is equal to :
A.
${{{3^{11}}} \over 2} + {2^{10}}$
B.
311 — 212
C.
2.311
D.
311
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
Let a1, a2, ..., an be a given A.P. whose
common difference is an integer and
Sn = a1 + a2 + .... + an. If a1 = 1, an = 300 and 15 $ \le $ n $ \le $ 50, then
the ordered pair (Sn-4, an–4) is equal to:
common difference is an integer and
Sn = a1 + a2 + .... + an. If a1 = 1, an = 300 and 15 $ \le $ n $ \le $ 50, then
the ordered pair (Sn-4, an–4) is equal to:
A.
(2480, 249)
B.
(2480, 248)
C.
(2490, 248)
D.
(2490, 249)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
The minimum value of 2sinx + 2cosx is :
A.
${2^{-1 + \sqrt 2 }}$
B.
${2^{1 - {1 \over {\sqrt 2 }}}}$
C.
${2^{1 - \sqrt 2 }}$
D.
${2^{-1 + {1 \over {\sqrt 2 }}}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
If 1+(1–22.1)+(1–42.3)+(1-62.5)+......+(1-202.19)= $\alpha $ - 220$\beta $,
then an ordered pair $\left( {\alpha ,\beta } \right)$ is equal to:
then an ordered pair $\left( {\alpha ,\beta } \right)$ is equal to:
A.
(11, 103)
B.
(10, 103)
C.
(10, 97)
D.
(11, 97)