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Sequences and Series

JEE Mathematics · 201 questions · Page 3 of 21 · Click an option or "Show Solution" to reveal answer

Q21
The sum 3×1313+5×(13+23)12+22+7×(13+23+33)12+22+32+.....{{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
Correct Answer
Option B
Solution
Tr=(2r+1)(13+23+33+......+r3)12+22+32+......+r2{T_r} = {{(2r + 1)({1^3} + {2^3} + {3^3} + ...... + {r^3})} \over {{1^2} + {2^2} + {3^2} + ...... + {r^2}}}
Tr=(2r+1)(r(r+1)2)2×6r(r+1)(2r+1){T_r} = (2r + 1){\left( {{{r(r + 1)} \over 2}} \right)^2} \times {6 \over {r(r + 1)(2r + 1)}}
Tr=3r(r+1)2{T_r} = {{3r(r + 1)} \over 2}

Now,

S=∑r=110Tr=32∑r=110(r2+r)S = \sum\limits_{r = 1}^{10} {{T_r}} = {3 \over 2}\sum\limits_{r = 1}^{10} {({r^2} + r)}
⇒32{10×(10+1)(2×10+1)6+10×112}\Rightarrow {3 \over 2}\left\{ {{{10 \times (10 + 1)(2 \times 10 + 1)} \over 6} + {{10 \times 11} \over 2}} \right\}
⇒32{10×11×216+5×11}\Rightarrow {3 \over 2}\left\{ {{{10 \times 11 \times 21} \over 6} + 5 \times 11} \right\}
⇒32×5×11×8=660\Rightarrow {3 \over 2} \times 5 \times 11 \times 8 = 660
Q22
The product 214.4116.8148.161128{2^{{1 \over 4}}}{.4^{{1 \over {16}}}}{.8^{{1 \over {48}}}}{.16^{{1 \over {128}}}} ... to ∞\infty is equal to :
A 214{2^{{1 \over 4}}}
B 212{2^{{1 \over 2}}}
C 1
D 2
Correct Answer
Option B
Solution
214.4116.8148.161128{2^{{1 \over 4}}}{.4^{{1 \over {16}}}}{.8^{{1 \over {48}}}}{.16^{{1 \over {128}}}}

... =

214+216+348+...∞{2^{{1 \over 4} + {2 \over {16}} + {3 \over {48}} + ...\infty }}

=

214+18+116+...∞{2^{{1 \over 4} + {1 \over 8} + {1 \over {16}} + ...\infty }}

=

2(141−12){2^{\left( {{{{1 \over 4}} \over {1 - {1 \over 2}}}} \right)}}

=

212{2^{{1 \over 2}}}
Q23
The minimum value of 2sinx + 2cosx is :
A 2−1+2{2^{-1 + \sqrt 2 }}
B 21−12{2^{1 - {1 \over {\sqrt 2 }}}}
C 21−2{2^{1 - \sqrt 2 }}
D 2−1+12{2^{-1 + {1 \over {\sqrt 2 }}}}
Correct Answer
Option B
Solution

Using AM ≥\ge GM

⇒2sin⁡ x+2cos⁡ x2≥2sin⁡x.2cos⁡x\Rightarrow {{{2^{\sin \,x}} + {2^{\cos \,x}}} \over 2} \ge \sqrt {{2^{\sin x}}{{.2}^{\cos x}}}
⇒2sin⁡x+2cos⁡x≥21+(sin⁡x+cos⁡x2)\Rightarrow {2^{\sin x}} + {2^{\cos x}} \ge {2^{1 + \left( {{{\sin x + \cos x} \over 2}} \right)}}
⇒min⁡(2sin⁡x+2cos⁡x)=21−12\Rightarrow \min ({2^{\sin x}} + {2^{\cos x}}) = {2^{1 - {1 \over {\sqrt 2 }}}}

As we know range of sin x + cos x is :

−2- \sqrt 2

≤\le sin x + cos x ≤\le

2\sqrt 2

. So Minimum value of sin x + cos x =

−2- \sqrt 2
Q24
If the sum of the first 15 terms of the series (34)3+(112)3+(214)3+33+(334)3+....{\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
Correct Answer
Option C
Solution

S =

(34)3+(64)3+(94)3+(124)3+ ........15{\left( {{3 \over 4}} \right)^3} + {\left( {{6 \over 4}} \right)^3} + {\left( {{9 \over 4}} \right)^3} + {\left( {{{12} \over 4}} \right)^3} + \,........15

term =

2764{{27} \over {64}}
∑r=115r3\sum\limits_{r = 1}^{15} {{r^3}}

=

2764.[15(15+1)2]2{{27} \over {64.}}{\left[ {{{15\left( {15 + 1} \right)} \over 2}} \right]^2}

= 225 K (Given in question) K = 27

Q25
If 1+(1–22.1)+(1–42.3)+(1-62.5)+......+(1-202.19)= α\alpha - 220β\beta , then an ordered pair (α,β)\left( {\alpha ,\beta } \right) is equal to:
A (11, 103)
B (10, 103)
C (10, 97)
D (11, 97)
Correct Answer
Option A
Solution
1+(1−22.1)+(1−42.3)+(1−62.5)+....(1−202.19)1 + (1 - {2^2}.1) + (1 - {4^2}.3) + (1 - {6^2}.5) + ....(1 - {20^2}.19)
S=1+∑r=110[1−(2r)2(2r−1)]=1+∑r=110(1−8r3+4r2)=1+10−∑r=110(8r3−4r2)S = 1 + \sum\limits_{r = 1}^{10} {\left[ {1 - {{(2r)}^2}(2r - 1)} \right] = 1 + \sum\limits_{r = 1}^{10} {\left( {1 - 8{r^3} + 4{r^2}} \right)} = 1 + 10 - } \sum\limits_{r = 1}^{10} {\left( {8{r^3} - 4{r^2}} \right)}
=11−8(10×112)2+4×(10×11×216)= 11 - 8{\left( {{{10 \times 11} \over 2}} \right)^2} + 4 \times \left( {{{10 \times 11 \times 21} \over 6}} \right)
=11−2×(110)2+4×55×7= 11 - 2 \times {(110)^2} + 4 \times 55 \times 7
=11−220(110−7)= 11 - 220(110 - 7)
=11−220×103=α−220β= 11 - 220 \times 103 = \alpha - 220\beta
⇒α=11,β=103\Rightarrow \alpha = 11,\beta = 103
Q26
Let an be the nth term of a G.P. of positive terms. ∑n=1100a2n+1=200\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200} and ∑n=1100a2n=100\sum\limits_{n = 1}^{100} {{a_{2n}} = 100} , then ∑n=1200an\sum\limits_{n = 1}^{200} {{a_n}} is equal to :
A 150
B 175
C 225
D 300
Correct Answer
Option A
Solution
∑n=1100a2n+1=200\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200}

⇒\Rightarrow a3 + a5 + a7 + .... + a201 = 200 ⇒\Rightarrow

ar2(r200−1)(r2−1)a{r^2}{{\left( {{r^{200}} - 1} \right)} \over {\left( {{r^2} - 1} \right)}}

= 200 ....(1)

∑n=1100a2n=100\sum\limits_{n = 1}^{100} {{a_{2n}} = 100}

⇒\Rightarrow a2 + a4 + a6 + ... + a200 = 100 ⇒\Rightarrow

ar(r200−1)(r2−1)ar{{\left( {{r^{200}} - 1} \right)} \over {\left( {{r^2} - 1} \right)}}

= 100 ....(2) dividing (1) by (2) we get, r = 2 adding both (1) and (2), we get a2 + a3 + a4 + a5 + ..... + a201 = 300 ⇒\Rightarrow r(a1 + a2 + ..... + a200) = 300 ⇒\Rightarrow a1 + a2 + ..... + a200 =

300r{{300} \over r}

⇒\Rightarrow

∑n=1200an\sum\limits_{n = 1}^{200} {{a_n}}

=

3002{{300} \over 2}

= 150

Q27
In an increasing geometric series, the sum of the second and the sixth term is 252{{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
Correct Answer
Option D
Solution

a, ar, ar2, .....

T2+T6=252⇒ar(1+r4)=252{T_2} + {T_6} = {{25} \over 2} \Rightarrow ar(1 + {r^4}) = {{25} \over 2}
a2r2(1+r4)2=6254{a^2}{r^2}{(1 + {r^4})^2} = {{625} \over 4}

.... (1)

T3.T5=25⇒(ar2)(ar4)=25{T_3}.{T_5} = 25 \Rightarrow (a{r^2})(a{r^4}) = 25
a2r6=25{a^2}{r^6} = 25

.....(2) On dividing (1) by (2)

(1+r4)2r4=254{{{{(1 + {r^4})}^2}} \over {{r^4}}} = {{25} \over 4}
4r8−14r4+4=04{r^8} - 14{r^4} + 4 = 0
(4r4−1)(r4−4)=0(4{r^4} - 1)({r^4} - 4) = 0
r4=14,4⇒r4=4{r^4} = {1 \over 4},4 \Rightarrow {r^4} = 4

(an increasing geometric series)

a2r6=25⇒(ar3)2=25{a^2}{r^6} = 25 \Rightarrow {(a{r^3})^2} = 25
T4+T6+T8=ar3+ar5+ar7{T_4} + {T_6} + {T_8} = a{r^3} + a{r^5} + a{r^7}
=ar3(1+r2+r4)= a{r^3}(1 + {r^2} + {r^4})
=5(1+2+4)=35= 5(1 + 2 + 4) = 35
Q28
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 )
Correct Answer
Option D
Solution

Let three terms of G.P. are

ar{a \over r}

, a, ar ∴\therefore

a(1r+1+r)a\left( {{1 \over r} + 1 + r} \right)

= S ...(1) and a3 = 27 ⇒\Rightarrow a = 3 ∴\therefore

3(1r+1+r)3\left( {{1 \over r} + 1 + r} \right)

= S ⇒\Rightarrow

1r+r=S3−1{{1 \over r} + r = {S \over 3} - 1}

⇒\Rightarrow As

1r+r≥2{{1 \over r} + r \ge 2}

or

1r+r≤−2{{1 \over r} + r \le - 2}

∴\therefore

S3−1≥2{{S \over 3} - 1 \ge 2}

or

S3−1≤−2{{S \over 3} - 1 \le - 2}

⇒\Rightarrow

S3≥3{{S \over 3} \ge 3}

or

S3≤−1{{S \over 3} \le - 1}

⇒\Rightarrow S ≥\ge 9 or S≤\le -3 ∴\therefore S

∈\in

(-∝\propto, -3] ∪\cup [9, ∞\infty)

Q29
The sum of the infinite series 1+23+732+1233+1734+2235+......1 + {2 \over 3} + {7 \over {{3^2}}} + {{12} \over {{3^3}}} + {{17} \over {{3^4}}} + {{22} \over {{3^5}}} + ...... is equal to :
A 94{9 \over 4}
B 134{13 \over 4}
C 154{15 \over 4}
D 114{11 \over 4}
Correct Answer
Option B
Solution
S=1+23+732+1233+1734+....S = 1 + {2 \over 3} + {7 \over {{3^2}}} + {{12} \over {{3^3}}} + {{17} \over {{3^4}}} + ....
S3=13+232+733+1234+....{S \over 3} = {1 \over 3} + {2 \over {{3^2}}} + {7 \over {{3^3}}} + {{12} \over {{3^4}}} + ....
2S=1+13+532+533+534+....2S = 1 + {1 \over 3} + {5 \over {{3^2}}} + {5 \over {{3^3}}} + {5 \over {{3^4}}} + ....

+ up to infinite terms

2S3=43+53{1/31−13}=56+43=136{{2S} \over 3} = {4 \over 3} + {5 \over 3}\left\{ {{{1/3} \over {1 - {1 \over 3}}}} \right\} = {5 \over 6} + {4 \over 3} = {{13} \over 6}

⇒\Rightarrow S =

134{13 \over 4}
Q30
132−1+152−1+172−1+....+1(201)2−1{1 \over {{3^2} - 1}} + {1 \over {{5^2} - 1}} + {1 \over {{7^2} - 1}} + .... + {1 \over {{{(201)}^2} - 1}} is equal to
A 101404{{101} \over {404}}
B 25101{{25} \over {101}}
C 101408{{101} \over {408}}
D 99400{{99} \over {400}}
Correct Answer
Option B
Solution
S=∑r=11001(2n+1)2−1S = \sum\limits_{r = 1}^{100} {{1 \over {{{(2n + 1)}^2} - 1}}}
=∑r=11001(2n+1+1)(2n+1−1)= \sum\limits_{r = 1}^{100} {{1 \over {(2n + 1 + 1)(2n + 1 - 1)}}}
=∑r=110012n(2n+2)= \sum\limits_{r = 1}^{100} {{1 \over {2n(2n + 2)}}}
=14∑r=11001n(n+1)= {1 \over 4}\sum\limits_{r = 1}^{100} {{1 \over {n(n + 1)}}}
=14∑r=1100(n+1)−nn(n+1)= {1 \over 4}\sum\limits_{r = 1}^{100} {{{(n + 1) - n} \over {n(n + 1)}}}
=14∑r=1100(1n−1n+1)= {1 \over 4}\sum\limits_{r = 1}^{100} {\left( {{1 \over n} - {1 \over {n + 1}}} \right)}
S=14((1−12)+(12−13)+(13−14)+......+(1100−1101))S = {1 \over 4}\left( {\left( {1 - {1 \over 2}} \right) + \left( {{1 \over 2} - {1 \over 3}} \right) + \left( {{1 \over 3} - {1 \over 4}} \right) + ...... + \left( {{1 \over {100}} - {1 \over {101}}} \right)} \right)

∴\therefore

S=14[100101]=25101S = {1 \over 4}\left[ {{{100} \over {101}}} \right] = {{25} \over {101}}
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