Sequences and Series
426 Questions
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)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
If the sum of the series
20 + 19${3 \over 5}$ + 19${1 \over 5}$ + 18${4 \over 5}$ + ...
upto nth term is 488 and the nth term is negative, then :
20 + 19${3 \over 5}$ + 19${1 \over 5}$ + 18${4 \over 5}$ + ...
upto nth term is 488 and the nth term is negative, then :
A.
n = 41
B.
n = 60
C.
nth term is –4
D.
nth term is -4${2 \over 5}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
If the first term of an A.P. is 3 and the sum of
its first 25 terms is equal to the sum of its next
15 terms, then the common difference of this
A.P. is :
A.
${1 \over 4}$
B.
${1 \over 5}$
C.
${1 \over 7}$
D.
${1 \over 6}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
Let S be the sum of the first 9 terms of the
series :
{x + k$a$} + {x2 + (k + 2)$a$} + {x3 + (k + 4)$a$}
+ {x4 + (k + 6)$a$} + .... where a $ \ne $ 0 and x $ \ne $ 1.
If S = ${{{x^{10}} - x + 45a\left( {x - 1} \right)} \over {x - 1}}$, then k is equal to :
{x + k$a$} + {x2 + (k + 2)$a$} + {x3 + (k + 4)$a$}
+ {x4 + (k + 6)$a$} + .... where a $ \ne $ 0 and x $ \ne $ 1.
If S = ${{{x^{10}} - x + 45a\left( {x - 1} \right)} \over {x - 1}}$, then k is equal to :
A.
-3
B.
1
C.
-5
D.
3
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
If the sum of first 11 terms of an A.P.,
a1, a2, a3, .... is 0 (a $ \ne $ 0), then the sum of the A.P.,
a1 , a3 , a5 ,....., a23 is ka1 , where k is equal to :
a1, a2, a3, .... is 0 (a $ \ne $ 0), then the sum of the A.P.,
a1 , a3 , a5 ,....., a23 is ka1 , where k is equal to :
A.
${{121} \over {10}}$
B.
-${{121} \over {10}}$
C.
${{72} \over 5}$
D.
-${{72} \over 5}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
The sum of the first three terms of a G.P. is S and
their product is 27. Then all such S lie in :
A.
[-3, $\infty $)
B.
(-$ \propto $, 9]
C.
(-$ \propto $, -9] $ \cup $ [-3, $\infty $)
D.
(-$ \propto $, -3] $ \cup $ [9, $\infty $)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
If |x| < 1, |y| < 1 and x $ \ne $ y, then the sum to infinity
of the following series
(x + y) + (x2+xy+y2) + (x3+x2y + xy2+y3) + ....
(x + y) + (x2+xy+y2) + (x3+x2y + xy2+y3) + ....
A.
${{x + y - xy} \over {\left( {1 + x} \right)\left( {1 + y} \right)}}$
B.
${{x + y - xy} \over {\left( {1 - x} \right)\left( {1 - y} \right)}}$
C.
${{x + y + xy} \over {\left( {1 + x} \right)\left( {1 + y} \right)}}$
D.
${{x + y + xy} \over {\left( {1 - x} \right)\left( {1 - y} \right)}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Evening Slot
Let an be the nth term of a G.P. of positive terms.
$\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200} $ and $\sum\limits_{n = 1}^{100} {{a_{2n}} = 100} $,
then $\sum\limits_{n = 1}^{200} {{a_n}} $ is equal to :
$\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200} $ and $\sum\limits_{n = 1}^{100} {{a_{2n}} = 100} $,
then $\sum\limits_{n = 1}^{200} {{a_n}} $ is equal to :
A.
150
B.
175
C.
225
D.
300
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
The product ${2^{{1 \over 4}}}{.4^{{1 \over {16}}}}{.8^{{1 \over {48}}}}{.16^{{1 \over {128}}}}$ ... to $\infty $ is equal
to :
A.
${2^{{1 \over 4}}}$
B.
${2^{{1 \over 2}}}$
C.
1
D.
2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
If the 10th term of an A.P. is ${1 \over {20}}$ and its 20th term
is ${1 \over {10}}$, then the sum of its first 200 terms is
A.
100
B.
$100{1 \over 2}$
C.
$50{1 \over 4}$
D.
50
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
Let ƒ : R $ \to $ R be such that for all
x $ \in $ R
(21+x + 21–x), ƒ(x) and (3x + 3–x) are in A.P.,
then the minimum value of ƒ(x) is
(21+x + 21–x), ƒ(x) and (3x + 3–x) are in A.P.,
then the minimum value of ƒ(x) is
A.
2
B.
0
C.
3
D.
4
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
If the sum of the first 40 terms of the series,
3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + ..... is (102)m, then m is equal to :
3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + ..... is (102)m, then m is equal to :
A.
20
B.
5
C.
10
D.
25
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
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\limits_{i = 1}^9 {{a_i}} = 4\lambda $, then $\lambda $ is equal to:
${a_1}$ < 0, ${a_1}$ + ${a_2}$ = 4 and ${a_3}$ + ${a_4}$ = 16.
If $\sum\limits_{i = 1}^9 {{a_i}} = 4\lambda $, then $\lambda $ is equal to:
A.
171
B.
-171
C.
-513
D.
${{511} \over 3}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
Five numbers are in A.P. whose sum is 25 and product is 2520. If one of these five numbers is -${1 \over 2}$ , then the greatest number amongst them is:
A.
${{21} \over 2}$
B.
27
C.
7
D.
16
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Evening Slot
If m arithmetic means (A.Ms) and three
geometric means (G.Ms) are inserted between
3 and 243 such that 4th A.M. is equal to 2nd
G.M., then m is equal to _________ .
Correct Answer: 39
Explanation:
Given m arithmetic means (A.Ms) present between 3 and 243
$ \therefore $ Common difference, $d = {{b - a} \over {m + 1}} = {{240} \over {m + 1}}$
$ \therefore $ 4th A.M. = a + 4d
= 3 + 4 $ \times $ ${{240} \over {m + 1}}$
Also there are 3 G.M between 3 and 243
$ \therefore $ Common ratio (r) = ${\left( {{b \over a}} \right)^{{1 \over {n + 1}}}}$
where n = number of G.M inserted.
$ \therefore $ r = ${\left( {{{243} \over 3}} \right)^{{1 \over {3 + 1}}}} = 3$
Given,
4th A.M = 2nd G.M
$ \Rightarrow 3 + 4 \times {{240} \over {m + 1}} = 3{(3)^2}$
$ \Rightarrow {{960} \over {m + 1}} = 24$
$ \Rightarrow m = 39$
$ \therefore $ Common difference, $d = {{b - a} \over {m + 1}} = {{240} \over {m + 1}}$
$ \therefore $ 4th A.M. = a + 4d
= 3 + 4 $ \times $ ${{240} \over {m + 1}}$
Also there are 3 G.M between 3 and 243
$ \therefore $ Common ratio (r) = ${\left( {{b \over a}} \right)^{{1 \over {n + 1}}}}$
where n = number of G.M inserted.
$ \therefore $ r = ${\left( {{{243} \over 3}} \right)^{{1 \over {3 + 1}}}} = 3$
Given,
4th A.M = 2nd G.M
$ \Rightarrow 3 + 4 \times {{240} \over {m + 1}} = 3{(3)^2}$
$ \Rightarrow {{960} \over {m + 1}} = 24$
$ \Rightarrow m = 39$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Morning Slot
The value of ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$ is equal to ______.
Correct Answer: 4
Explanation:
Given, ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$
As sum of GP upto infinity = ${a \over {1 - r}}$
$ \therefore $ ${1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....\infty $ = ${{{1 \over 3}} \over {1 - {1 \over 3}}}$ = ${1 \over 2}$
$ \therefore $ ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$
= ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left( {{{16} \over {100}}} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left( {{4 \over {10}}} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left[ {{{\left( {{{10} \over 4}} \right)}^{ - 2}}} \right]^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left[ {{{\left( {2.5} \right)}^{ - 2}}} \right]^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${{{\left( {2.5} \right)}^{ - 2{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}}$
= ${{{\left( {{1 \over 2}} \right)}^{ - 2}}}$ = 4
As sum of GP upto infinity = ${a \over {1 - r}}$
$ \therefore $ ${1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....\infty $ = ${{{1 \over 3}} \over {1 - {1 \over 3}}}$ = ${1 \over 2}$
$ \therefore $ ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$
= ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left( {{{16} \over {100}}} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left( {{4 \over {10}}} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left[ {{{\left( {{{10} \over 4}} \right)}^{ - 2}}} \right]^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left[ {{{\left( {2.5} \right)}^{ - 2}}} \right]^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${{{\left( {2.5} \right)}^{ - 2{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}}$
= ${{{\left( {{1 \over 2}} \right)}^{ - 2}}}$ = 4
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Evening Slot
The number of terms common to the two A.P.'s
3, 7, 11, ....., 407 and 2, 9, 16, ....., 709 is ______.
Correct Answer: 14
Explanation:
First A.P. is 3, 7, 11, 15, 19, 23, ..... 407
d1 = 4
Second A.P. is 2, 9, 16, 23, ..... 709
d2 = 7
First common term = 23
Common difference of new A.P using the common terms of the two given A.P's is d = L.C.M. (4, 7) = 28
Last term $ \le $ 407
$ \Rightarrow $ 23 + (n – 1) (28) $ \le $ 407
$ \Rightarrow $ n $ \le $ 14.7
$ \therefore $ n = 14
d1 = 4
Second A.P. is 2, 9, 16, 23, ..... 709
d2 = 7
First common term = 23
Common difference of new A.P using the common terms of the two given A.P's is d = L.C.M. (4, 7) = 28
Last term $ \le $ 407
$ \Rightarrow $ 23 + (n – 1) (28) $ \le $ 407
$ \Rightarrow $ n $ \le $ 14.7
$ \therefore $ n = 14
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 8th January Evening Slot
The sum, $\sum\limits_{n = 1}^7 {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 4}} $ is equal to
________.
Correct Answer: 504
Explanation:
$\sum\limits_{n = 1}^7 {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 4}} $
= ${1 \over 4}\sum\limits_{n = 1}^7 {\left( {2{n^3} + 3{n^2} + n} \right)} $
= ${1 \over 2}\sum\limits_{n = 1}^7 {{n^3}} $ + ${3 \over 4}\sum\limits_{n = 1}^7 {{n^2}} $ + ${1 \over 4}\sum\limits_{n = 1}^7 n $
= ${1 \over 2}{\left( {{{7\left( {7 + 1} \right)} \over 2}} \right)^2}$ + ${3 \over 4}\left( {{{7\left( {7 + 1} \right)\left( {14 + 1} \right)} \over 6}} \right)$ + ${1 \over 4}{{7\left( 8 \right)} \over 2}$
= (49)(8) + (15$ \times $7) + (7)
= 392 + 105 + 7 = 504
= ${1 \over 4}\sum\limits_{n = 1}^7 {\left( {2{n^3} + 3{n^2} + n} \right)} $
= ${1 \over 2}\sum\limits_{n = 1}^7 {{n^3}} $ + ${3 \over 4}\sum\limits_{n = 1}^7 {{n^2}} $ + ${1 \over 4}\sum\limits_{n = 1}^7 n $
= ${1 \over 2}{\left( {{{7\left( {7 + 1} \right)} \over 2}} \right)^2}$ + ${3 \over 4}\left( {{{7\left( {7 + 1} \right)\left( {14 + 1} \right)} \over 6}} \right)$ + ${1 \over 4}{{7\left( 8 \right)} \over 2}$
= (49)(8) + (15$ \times $7) + (7)
= 392 + 105 + 7 = 504
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 8th January Morning Slot
The sum $\sum\limits_{k = 1}^{20} {\left( {1 + 2 + 3 + ... + k} \right)} $ is :
Correct Answer: 1540
Explanation:
$\sum\limits_{k = 1}^{20} {\left( {1 + 2 + 3 + ... + k} \right)} $
= $\sum\limits_{k = 1}^{20} {{{k\left( {k + 1} \right)} \over 2}} $
= $\sum\limits_{k = 1}^{20} {{{{k^2}} \over 2}} + \sum\limits_{k = 1}^{20} {{k \over 2}} $
= ${1 \over 2} \times {{20 \times 21 \times 41} \over 6} + {1 \over 2} \times {{20 \times 21} \over 2}$
= 1540
= $\sum\limits_{k = 1}^{20} {{{k\left( {k + 1} \right)} \over 2}} $
= $\sum\limits_{k = 1}^{20} {{{{k^2}} \over 2}} + \sum\limits_{k = 1}^{20} {{k \over 2}} $
= ${1 \over 2} \times {{20 \times 21 \times 41} \over 6} + {1 \over 2} \times {{20 \times 21} \over 2}$
= 1540
2020
JEE Advanced
Numerical
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 ...........
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 have
${{{3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}}} \over 3} \ge {({3^{{y_1}}}\,.\,{3^{{y_2}}}\,.\,{3^{{y_3}}})^{{1 \over 3}}}$
$ \Rightarrow {3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}} \ge 3{({3^{{y_1}}}\,.\,{3^{{y_2}}}\,.\,{3^{{y_3}}})^{{1 \over 3}}}$
On applying logarithm with base '3', we get
${\log _3}({3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}}) \ge \left[ {1 + {1 \over 3}({y_1} + {y_2} + {y_3}} \right.)]$
= 1 + 3
= 4
{$ \because $ ${{y_1} + {y_2} + {y_3}}$ = 9}
$ \therefore $ m = 4
Now, for positive real numbers x1, x2 and x3 according to AM-GM inequality, we have
${{{x_1} + {x_2} + {x_3}} \over 3} \ge {({x_1}{x_2}{x_3})^{{1 \over 3}}}$
On applying logarithm with base '3', we get
${\log _3}\left( {{{{x_1} + {x_2} + {x_3}} \over 3}} \right) \ge {1 \over 3}$$({\log _3}{x_1} + {\log _3}{x_2} + {\log _3}{x_3})$
$ \Rightarrow $ $1 \ge {1 \over 3}\left( {{{\log }_3}{x_1} + {{\log }_3}{x_2} + {{\log }_3}{x_3}} \right)$
{$ \because $ x1 + x2 + x3 = 9}
$ \therefore $ M = 3
Now, ${\log _2}({m^3}) + {\log _3}({M^2})$
$ = 3lo{g_2}(4) + 2lo{g_2}(3)$ = 6 + 2 = 8
${{{3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}}} \over 3} \ge {({3^{{y_1}}}\,.\,{3^{{y_2}}}\,.\,{3^{{y_3}}})^{{1 \over 3}}}$
$ \Rightarrow {3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}} \ge 3{({3^{{y_1}}}\,.\,{3^{{y_2}}}\,.\,{3^{{y_3}}})^{{1 \over 3}}}$
On applying logarithm with base '3', we get
${\log _3}({3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}}) \ge \left[ {1 + {1 \over 3}({y_1} + {y_2} + {y_3}} \right.)]$
= 1 + 3
= 4
{$ \because $ ${{y_1} + {y_2} + {y_3}}$ = 9}
$ \therefore $ m = 4
Now, for positive real numbers x1, x2 and x3 according to AM-GM inequality, we have
${{{x_1} + {x_2} + {x_3}} \over 3} \ge {({x_1}{x_2}{x_3})^{{1 \over 3}}}$
On applying logarithm with base '3', we get
${\log _3}\left( {{{{x_1} + {x_2} + {x_3}} \over 3}} \right) \ge {1 \over 3}$$({\log _3}{x_1} + {\log _3}{x_2} + {\log _3}{x_3})$
$ \Rightarrow $ $1 \ge {1 \over 3}\left( {{{\log }_3}{x_1} + {{\log }_3}{x_2} + {{\log }_3}{x_3}} \right)$
{$ \because $ x1 + x2 + x3 = 9}
$ \therefore $ M = 3
Now, ${\log _2}({m^3}) + {\log _3}({M^2})$
$ = 3lo{g_2}(4) + 2lo{g_2}(3)$ = 6 + 2 = 8
2020
JEE Advanced
Numerical
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 ...........
Correct Answer: 1
Explanation:
Given arithmetic progression of positive integers terms a1, a2, a3, ..... having common difference '2' and geometric progression of positive integers
terms b1, b2, b3, .... having common ratio '2' with a1 = b1 = c, such that
2(a1 + a2 + a3 + ... + an) = b1 + b2 + b3 + ... + bn
$ \Rightarrow 2 \times {n \over 2}[2C + (n - 1)2] = C\left( {{{{2^n} - 1} \over {2 - 1}}} \right)$
$ \Rightarrow 2nC + 2{n^2} - 2n = {2^n}.C - C$
$ \Rightarrow C[{2^n} - 2n - 1] = 2{n^2} - 2n$
$ \because $ $C \in N \Rightarrow 2{n^2} - 2n \ge {2^n} - 2n - 1$
$ \Rightarrow 2{n^2} + 1 \ge {2^n} \Rightarrow n \le 6$
and, also C > 0 $ \Rightarrow $ n > 2
$ \therefore $ The possible values of n are 3, 4, 5, 6
So, at $n = 3,\,C = {{(2 \times 9) - 6} \over {8 - 6 - 1}} = 12$
at, $n = 4,\,C = {{32 - 8} \over {16 - 8 - 1}} = {{24} \over 9} = {8 \over 3} \notin N$
at, $n = 5,\,C = {{50 - 10} \over {32 - 10 - 1}} = {{40} \over {21}} \notin N$
and at, $n = 6,\,C = {{72 - 12} \over {64 - 12 - 1}} = {{60} \over {51}} \notin N$
$ \therefore $ The required value of C = 12 for n = 3
so number of possible value of C is 1
terms b1, b2, b3, .... having common ratio '2' with a1 = b1 = c, such that
2(a1 + a2 + a3 + ... + an) = b1 + b2 + b3 + ... + bn
$ \Rightarrow 2 \times {n \over 2}[2C + (n - 1)2] = C\left( {{{{2^n} - 1} \over {2 - 1}}} \right)$
$ \Rightarrow 2nC + 2{n^2} - 2n = {2^n}.C - C$
$ \Rightarrow C[{2^n} - 2n - 1] = 2{n^2} - 2n$
$ \because $ $C \in N \Rightarrow 2{n^2} - 2n \ge {2^n} - 2n - 1$
$ \Rightarrow 2{n^2} + 1 \ge {2^n} \Rightarrow n \le 6$
and, also C > 0 $ \Rightarrow $ n > 2
$ \therefore $ The possible values of n are 3, 4, 5, 6
So, at $n = 3,\,C = {{(2 \times 9) - 6} \over {8 - 6 - 1}} = 12$
at, $n = 4,\,C = {{32 - 8} \over {16 - 8 - 1}} = {{24} \over 9} = {8 \over 3} \notin N$
at, $n = 5,\,C = {{50 - 10} \over {32 - 10 - 1}} = {{40} \over {21}} \notin N$
and at, $n = 6,\,C = {{72 - 12} \over {64 - 12 - 1}} = {{60} \over {51}} \notin N$
$ \therefore $ The required value of C = 12 for n = 3
so number of possible value of C is 1
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
Let $f(n)=A(-2)^n+B(-3)^n \forall A, B \in \mathbf{R}$ and $n \in \mathbf{N}-\{1,2\}$. If $f(n)+a f(n-1)+b f(n-2)=0$, then $(a+b)(b-a)=$
A.
0
B.
5
C.
7
D.
11
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
If $1+\frac{\cos \theta}{2}+\frac{\cos 2 \theta}{4}+\frac{\cos 3 \theta}{8}+\ldots \ldots=\frac{a-2 \cos \theta}{5+b \cos \theta}$ for some $a, b \in \mathbf{R}$, then $(a-b)^2=$
A.
0
B.
64
C.
36
D.
125
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Evening Shift
If $S_n$ is the sum of the first $n$ terms of the series $1^2+2 \times 2^2+3^2+2 \times 4^2+5^2+2 \times 6^2+\ldots \infty$, then, when $n$ is even $S_n=$
A.
$\frac{n(n+1)}{2}$
B.
$\frac{n^2(n+1)}{2}$
C.
$\frac{n(n+1)^2}{2}$
D.
$\frac{n^2(n+2)}{2}$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Evening Shift
If the roots of the equation, $8 x^3+6 p x^2+3 q x-27=0$ are in a geometric progression, then $q^2+9 p^2+6 p q+q / p=$
A.
-3
B.
-10
C.
6
D.
0
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Morning Shift
Let the greatest common divisor of $m, n$ be 1 . If $\frac{1}{1 \cdot 7}+\frac{1}{7 \cdot 13}+\frac{1}{13 \cdot 19}+\ldots \ldots$. upto 20 terms $=\frac{m}{n}$, then $5 m+2 n=$
A.
325
B.
330
C.
342
D.
337
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
If a1, a2, a3, ..... are in A.P. such that a1 + a7 + a16 = 40, then the sum of the first 15 terms of this A.P. is :
A.
120
B.
200
C.
150
D.
280
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
For x $\varepsilon $ R, let [x] denote the greatest integer $ \le $ x, then the sum of the series
$\left[ { - {1 \over 3}} \right] + \left[ { - {1 \over 3} - {1 \over {100}}} \right] + \left[ { - {1 \over 3} - {2 \over {100}}} \right] + .... + \left[ { - {1 \over 3} - {{99} \over {100}}} \right]$ is :
A.
- 153
B.
- 135
C.
- 133
D.
- 131
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
Let Sn denote the sum of the first n terms of an A.P. If S4 = 16 and S6= – 48, then S10 is equal to :
A.
- 320
B.
- 380
C.
- 460
D.
- 210
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
Let a1, a2, a3,......be an A.P. with a6 = 2. Then the common difference of this A.P., which maximises the
product a1a4a5, is :
A.
${3 \over 2}$
B.
${6 \over 5}$
C.
${8 \over 5}$
D.
${2 \over 3}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
The sum
$1 + {{{1^3} + {2^3}} \over {1 + 2}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 2 + 3}} + ...... + {{{1^3} + {2^3} + {3^3} + ... + {{15}^3}} \over {1 + 2 + 3 + ... + 15}}$$ - {1 \over 2}\left( {1 + 2 + 3 + ... + 15} \right)$ is equal to :
$1 + {{{1^3} + {2^3}} \over {1 + 2}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 2 + 3}} + ...... + {{{1^3} + {2^3} + {3^3} + ... + {{15}^3}} \over {1 + 2 + 3 + ... + 15}}$$ - {1 \over 2}\left( {1 + 2 + 3 + ... + 15} \right)$ is equal to :
A.
620
B.
1240
C.
1860
D.
660
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
Let $a$, b and c be in G.P. with common ratio r, where $a$ $ \ne $ 0 and 0 < r $ \le $ ${1 \over 2}$
. If 3$a$, 7b and 15c are the first three
terms of an A.P., then the 4th term of this A.P. is :
A.
$a$
B.
${7 \over 3}a$
C.
5$a$
D.
${2 \over 3}a$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
If a1, a2, a3, ............... an are in A.P. and a1 + a4 + a7 + ........... + a16 = 114, then a1 + a6 + a11 + a16 is equal to :
A.
38
B.
98
C.
76
D.
64
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
The sum
${{3 \times {1^3}} \over {{1^3}}} + {{5 \times ({1^3} + {2^3})} \over {{1^2} + {2^2}}} + {{7 \times \left( {{1^3} + {2^3} + {3^3}} \right)} \over {{1^2} + {2^2} + {3^2}}} + .....$ upto 10 terms is:
${{3 \times {1^3}} \over {{1^3}}} + {{5 \times ({1^3} + {2^3})} \over {{1^2} + {2^2}}} + {{7 \times \left( {{1^3} + {2^3} + {3^3}} \right)} \over {{1^2} + {2^2} + {3^2}}} + .....$ upto 10 terms is:
A.
600
B.
660
C.
680
D.
620
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
Some identical balls are arranged in rows to form
an equilateral triangle. The first row consists of one
ball, the second row consists of two balls and so
on. If 99 more identical balls are addded to the total
number of balls used in forming the equilaterial
triangle, then all these balls can be arranged in a
square whose each side contains exactly 2 balls
less than the number of balls each side of the
triangle contains. Then the number of balls used to
form the equilateral triangle is :-
A.
262
B.
190
C.
157
D.
225
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
If the sum and product of the first three term in
an A.P. are 33 and 1155, respectively, then a value
of its 11th term is :-
A.
–25
B.
–36
C.
25
D.
–35
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 +....
upto 11th term is :-
A.
945
B.
916
C.
915
D.
946
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
Let the sum of the first n terms of a non-constant
A.P., a1, a2, a3, ..... be $50n + {{n(n - 7)} \over 2}A$, where
A is a constant. If d is the common difference of
this A.P., then the ordered pair (d, a50) is equal to
A.
(A, 50+45A)
B.
(50, 50+45A)
C.
(A, 50+46A)
D.
(50, 50+46A)
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
If three distinct numbers a, b, c are in G.P. and the
equations ax2
+ 2bx + c = 0 and
dx2
+ 2ex + ƒ = 0 have a common root, then
which one of the following statements is
correct?
A.
$d \over a$, $e \over b$, $f \over c$ are in G.P.
B.
d, e, ƒ are in A.P
C.
d, e, ƒ are in G.P
D.
$d \over a$, $e \over b$, $f \over c$ are in A.P.
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
The sum
$\sum\limits_{k = 1}^{20} {k{1 \over {{2^k}}}} $ is equal to
A.
$2 - {11 \over {{2^{19}}}}$
B.
$2 - {3 \over {{2^{17}}}}$
C.
$1 - {11 \over {{2^{20}}}}$
D.
$2 - {21 \over {{2^{20}}}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Morning Slot
The sum of all natural numbers 'n' such that
100 < n < 200 and H.C.F. (91, n) > 1 is :
A.
3221
B.
3121
C.
3203
D.
3303
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
If sin4$\alpha $ + 4 cos4$\beta $ + 2 = 4$\sqrt 2 $ sin $\alpha $ cos $\beta $; $\alpha $, $\beta $ $ \in $ [0, $\pi $],
then cos($\alpha $ + $\beta $) $-$ cos($\alpha $ $-$ $\beta $) is equal to :
then cos($\alpha $ + $\beta $) $-$ cos($\alpha $ $-$ $\beta $) is equal to :
A.
$ - \sqrt 2 $
B.
0
C.
$-$ 1
D.
$\sqrt 2 $
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
If the sum of the first 15 terms of the series ${\left( {{3 \over 4}} \right)^3} + {\left( {1{1 \over 2}} \right)^3} + {\left( {2{1 \over 4}} \right)^3} + {3^3} + {\left( {3{3 \over 4}} \right)^3} + ....$ is equal to 225 k, then k
is equal to :
A.
9
B.
108
C.
27
D.
54
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
If nC4, nC5 and nC6 are in A.P., then n can be :
A.
11
B.
12
C.
9
D.
14
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Morning Slot
Let Sk = ${{1 + 2 + 3 + .... + k} \over k}.$ If $S_1^2 + S_2^2 + .....\, + S_{10}^2 = {5 \over {12}}$A, then A is equal to :
A.
283
B.
156
C.
301
D.
303
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Morning Slot
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 :
A.
36
B.
28
C.
32
D.
24
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Evening Slot
Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression ${{{x^m}{y^n}} \over {\left( {1 + {x^{2m}}} \right)\left( {1 + {y^{2n}}} \right)}}$ is :
A.
${1 \over 2}$
B.
${1 \over 4}$
C.
${{m + n} \over {6mn}}$
D.
1
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Evening Slot
If 19th term of a non-zero A.P. is zero, then its (49th term) : (29th term) is :
A.
2 : 1
B.
4 : 1
C.
1 : 3
D.
3 : 1