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Sequences and Series

JEE Mathematics · 201 questions · Page 5 of 21 · Click an option or "Show Solution" to reveal answer

Q41
If A=∑n=1∞1(3+(−1)n)nA = \sum\limits_{n = 1}^\infty {{1 \over {{{\left( {3 + {{( - 1)}^n}} \right)}^n}}}} and B=∑n=1∞(−1)n(3+(−1)n)nB = \sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{{\left( {3 + {{( - 1)}^n}} \right)}^n}}}} , then AB{A \over B} is equal to :
A 119{{11} \over 9}
B 1
C −-119{{11} \over 9}
D −-113{{11} \over 3}
Correct Answer
Option C
Solution
A=∑n=1∞1(3+(−1)n)nA = \sum\limits_{n = 1}^\infty {{1 \over {{{(3 + {{( - 1)}^n})}^n}}}}

and

B=∑n=1∞(−1)n(3+(−1)n)nB = \sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{{(3 + {{( - 1)}^n})}^n}}}}
A=12+142+123+144+A = {1 \over 2} + {1 \over {{4^2}}} + {1 \over {{2^3}}} + {1 \over {{4^4}}} +

........

B=−12+142−123+144+B = {{ - 1} \over 2} + {1 \over {{4^2}}} - {1 \over {{2^3}}} + {1 \over {{4^4}}} +

......

A=121−14+1161−116A = {{{1 \over 2}} \over {1 - {1 \over 4}}} + {{{1 \over {16}}} \over {1 - {1 \over {16}}}}

,

B=−121−14+1161−116B = {{ - {1 \over 2}} \over {1 - {1 \over 4}}} + {{{1 \over {16}}} \over {1 - {1 \over {16}}}}
A=1115A = {{11} \over {15}}

,

B=−915B = {{ - 9} \over {15}}

∴\therefore

AB=−119{A \over B} = {{ - 11} \over 9}
Q42
Let {an}n=0∞\{ {a_n}\} _{n = 0}^\infty be a sequence such that a0=a1=0{a_0} = {a_1} = 0 and an+2=2an+1−an+1{a_{n + 2}} = 2{a_{n + 1}} - {a_n} + 1 for all n ≥\ge 0. Then, ∑n=2∞an7n\sum\limits_{n = 2}^\infty {{{{a_n}} \over {{7^n}}}} is equal to:
A 6343{6 \over {343}}
B 7216{7 \over {216}}
C 8343{8 \over {343}}
D 49216{{49} \over {216}}
Correct Answer
Option B
Solution
an+2=2an+1−an+1{a_{n + 2}} = 2{a_{n + 1}} - {a_n} + 1

&

a0=a1=0{a_0} = {a_1} = 0
a2=2a1−a0+1=1{a_2} = 2{a_1} - {a_0} + 1 = 1
a3=2a2−a1+1=3{a_3} = 2{a_2} - {a_1} + 1 = 3
a4=2a3−a2+1=6{a_4} = 2{a_3} - {a_2} + 1 = 6
a5=2a4−a3+1=10{a_5} = 2{a_4} - {a_3} + 1 = 10
∑n=2∞an7n=a272+a373+a474+  ...\sum\limits_{n = 2}^\infty {{{{a_n}} \over {{7^n}}} = {{{a_2}} \over {{7^2}}} + {{{a_3}} \over {{7^3}}} + {{{a_4}} \over {{7^4}}} + \,\,...}
s=172+373+674+1075+  ...s = {1 \over {{7^2}}} + {3 \over {{7^3}}} + {6 \over {{7^4}}} + {{10} \over {{7^5}}} + \,\,...
17s=173+374+675+  ...6s7=172+273+374+  ...{{{1 \over 7}s = {1 \over {{7^3}}} + {3 \over {{7^4}}} + {6 \over {{7^5}}} + \,\,...} \over {{{6s} \over 7} = {1 \over {{7^2}}} + {2 \over {{7^3}}} + {3 \over {{7^4}}} + \,\,...}}
6s49=         173+274+  ...36s49=172+173+174+  ...{{{{6s} \over {49}} = \,\,\,\,\,\,\,\,\,{1 \over {{7^3}}} + {2 \over {{7^4}}} + \,\,...} \over {{{36s} \over {49}} = {1 \over {{7^2}}} + {1 \over {{7^3}}} + {1 \over {{7^4}}} + \,\,...}}
36s49=1721−17{{36s} \over {49}} = {{{1 \over {{7^2}}}} \over {1 - {1 \over 7}}}
36s49=749×6{{36s} \over {49}} = {7 \over {49 \times 6}}
s=7216s = {7 \over {216}}
Q43
The sum 1 + 2 . 3 + 3 . 32 + ......... + 10 . 39 is equal to :
A 2 . 312+104{{2\,.\,{3^{12}} + 10} \over 4}
B 19 . 310+14{{19\,.\,{3^{10}} + 1} \over 4}
C 5 . 310−25\,.\,{3^{10}} - 2
D 9 . 310+12{{9\,.\,{3^{10}} + 1} \over 2}
Correct Answer
Option B
Solution

Let

S=1 . 30+2 . 31+3 . 32+  ......  +  10 . 39S = 1\,.\,{3^0} + 2\,.\,{3^1} + 3\,.\,{3^2} + \,\,......\,\, + \,\,10\,.\,{3^9}
3S=1 . 31+2 . 32+  ..........  +  10 . 3103S = 1\,.\,{3^1} + 2\,.\,{3^2} + \,\,..........\,\, + \,\,10\,.\,{3^{10}}

___________________________________________________________

−2S=(1 . 30+1 . 31+1 . 32+  ........  +  1 . 39)−10 . 310- 2S = (1\,.\,{3^0} + 1\,.\,{3^1} + 1\,.\,{3^2} + \,\,........\,\, + \,\,1\,.\,{3^9}) - 10\,.\,{3^{10}}
⇒S=12[10 . 310−310−1−3−1]\Rightarrow S = {1 \over 2}\left[ {10\,.\,{3^{10}} - {{{3^{10}} - 1} \over { - 3 - 1}}} \right]
⇒S=19 . 310+14\Rightarrow S = {{19\,.\,{3^{10}} + 1} \over 4}
Q44
If {ai}i=1n\{ {a_i}\} _{i = 1}^n, where n is an even integer, is an arithmetic progression with common difference 1, and ∑i=1nai=192, ∑i=1n/2a2i=120\sum\limits_{i = 1}^n {{a_i} = 192} ,\,\sum\limits_{i = 1}^{n/2} {{a_{2i}} = 120} , then n is equal to :
A 48
B 96
C 92
D 104
Correct Answer
Option B
Solution
∑i=1nai=192\sum\limits_{i = 1}^n {{a_i} = 192}

⇒\Rightarrow a1 + a2 + a3 + ...... + an = 192

⇒n2[a1+an]=192\Rightarrow {n \over 2}[{a_1} + {a_n}] = 192
⇒a1+an=384n\Rightarrow {a_1} + {a_n} = {{384} \over n}

..... (1) Now,

∑i=1n2a2i=120\sum\limits_{i = 1}^{{n \over 2}} {{a_{2i}} = 120}

⇒\Rightarrow a2 + a4 + a6 + ...... + an = 120 Here total

n2{n \over 2}

terms present. ∴\therefore

n22[a2+an]=120{{{n \over 2}} \over 2}[{a_2} + {a_n}] = 120
⇒n4[a1+1+an]=120\Rightarrow {n \over 4}[{a_1} + 1 + {a_n}] = 120
⇒a1+an+1=480n\Rightarrow {a_1} + {a_n} + 1 = {{480} \over n}

..... (2) Subtracting (1) from (2), we get

1=480n−384n1 = {{480} \over n} - {{384} \over n}
⇒1=96n\Rightarrow 1 = {{96} \over n}

⇒\Rightarrow n = 96

Q45
The value of 1+11+2+11+2+3+  ....  +  11+2+3+  .....  +  111 + {1 \over {1 + 2}} + {1 \over {1 + 2 + 3}} + \,\,....\,\, + \,\,{1 \over {1 + 2 + 3 + \,\,.....\,\, + \,\,11}} is equal to:
A 2011{{20} \over {11}}
B 116{{11} \over {6}}
C 241132{{241} \over {132}}
D 2111{{21} \over {11}}
Correct Answer
Option B
Solution

Given,

1+11+2+11+2+3 + ....  +  11+2+3+  ....  + 111 + {1 \over {1 + 2}} + {1 \over {1 + 2 + 3}}\, + \,....\,\, + \,\,{1 \over {1 + 2 + 3 + \,\,....\,\, + \,11}}

General term,

Tn=11+2+3 +  ....  +  n{T_n} = {1 \over {1 + 2 + 3\, + \,\,....\,\, + \,\,n}}
=1n(n+1)2= {1 \over {{{n(n + 1)} \over 2}}}
=2n(n+1)= {2 \over {n(n + 1)}}
=2[n+1−nn(n+1)]= 2\left[ {{{n + 1 - n} \over {n(n + 1)}}} \right]
=2[1n−1n+1]= 2\left[ {{1 \over n} - {1 \over {n + 1}}} \right]
t1=2[11−12]{t_1} = 2\left[ {{1 \over 1} - {1 \over 2}} \right]
t2=2[12−13]{t_2} = 2\left[ {{1 \over 2} - {1 \over 3}} \right]
t3=2[13−14]{t_3} = 2\left[ {{1 \over 3} - {1 \over 4}} \right]

⋮\vdots

tn=2[1n−1n+1]{t_n} = 2\left[ {{1 \over n} - {1 \over {n + 1}}} \right]

∴\therefore

Sn=t1+t2+t3 +  ....  +  tn{S_n} = {t_1} + {t_2} + {t_3}\, + \,\,....\,\, + \,\,{t_n}
=2[11−12+12−13+13−14 +  ...  +  1n−1n+1]= 2\left[ {{1 \over 1} - {1 \over 2} + {1 \over 2} - {1 \over 3} + {1 \over 3} - {1 \over 4}\, + \,\,...\,\, + \,\,{1 \over n} - {1 \over {n + 1}}} \right]
=2[1−1n+1]= 2\left[ {1 - {1 \over {n + 1}}} \right]
=2[n+1−1n+1]= 2\left[ {{{n + 1 - 1} \over {n + 1}}} \right]
=2nn+1= {{2n} \over {n + 1}}

∴\therefore

S11=2×1111+1=2212=116{S_{11}} = {{2 \times 11} \over {11 + 1}} = {{22} \over {12}} = {{11} \over 6}
Q46
The sum ∑n=1213(4n−1)(4n+3)\sum\limits_{n = 1}^{21} {{3 \over {(4n - 1)(4n + 3)}}} is equal to
A 787\dfrac{7}{87}
B 729\dfrac{7}{29}
C 1487\dfrac{14}{87}
D 2129\dfrac{21}{29}
Correct Answer
Option B
Solution
∑n=1213(4n−1)(4n+3)=34∑n=12114n−1−14n+3\sum\limits_{n = 1}^{21} {{3 \over {(4n - 1)(4n + 3)}} = {3 \over 4}\sum\limits_{n = 1}^{21} {{1 \over {4n - 1}} - {1 \over {4n + 3}}} }
=34[(13−17)+(17−111)+  ....  +  (183−187)]= {3 \over 4}\left[ {\left( {{1 \over 3} - {1 \over 7}} \right) + \left( {{1 \over 7} - {1 \over {11}}} \right) + \,\,....\,\, + \,\,\left( {{1 \over {83}} - {1 \over {87}}} \right)} \right]
=34[13−187]=34843.87=729= {3 \over 4}\left[ {{1 \over 3} - {1 \over {87}}} \right] = {3 \over 4}{{84} \over {3.87}} = {7 \over {29}}
Q47
Consider two G.Ps. 2, 22, 23, ..... and 4, 42, 43, .... of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is (2)2258{(2)^{{{225} \over 8}}}, then ∑k=1nk(n−k)\sum\limits_{k = 1}^n {k(n - k)} is equal to :
A 560
B 1540
C 1330
D 2600
Correct Answer
Option C
Solution

Given G.P's 2, 22, 23, .... 60 terms 4, 42, .... n terms Now, G.M

=22258= {2^{{{225} \over 8}}}
(2.22... 4.42...)160+n=22258{\left( {{{2.2}^2}...\,{{4.4}^2}...} \right)^{{1 \over {60 + n}}}} = {2^{{{225} \over 8}}}
(2n2+n+183060+n)=22258\left( {{2^{{{{n^2} + n + 1830} \over {60 + n}}}}} \right) = {2^{{{225} \over 8}}}
⇒n2+n+183060+n=2258\Rightarrow {{{n^2} + n + 1830} \over {60 + n}} = {{225} \over 8}
⇒8n2−217n+1140=0\Rightarrow 8{n^2} - 217n + 1140 = 0
n=578, 20, n = {{57} \over 8},\,20,\,

so

n=20n = 20

∴\therefore

∑k=120k(20−k)=20×20×212−20×21×416\sum\limits_{k = 1}^{20} {k(20 - k) = 20 \times {{20 \times 21} \over 2} - {{20 \times 21 \times 41} \over 6}}
=20×212[20−413]=1330= {{20 \times 21} \over 2}\left[ {20 - {{41} \over 3}} \right] = 1330
Q48
Suppose a1,a2,…,ana_{1}, a_{2}, \ldots, a_{n}, .. be an arithmetic progression of natural numbers. If the ratio of the sum of first five terms to the sum of first nine terms of the progression is 5:175: 17 and , $$110
A 290
B 380
C 460
D 510
Correct Answer
Option B
Solution

∵\because a1, a2, .... an be an A.P of natural numbers and

S5S9=517⇒52[2a1+4d]92[2a1+8d]=517{{{S_5}} \over {{S_9}}} = {5 \over {17}} \Rightarrow {{{5 \over 2}[2{a_1} + 4d]} \over {{9 \over 2}[2{a_1} + 8d]}} = {5 \over {17}}
⇒34a1+68d=18a1+72d\Rightarrow 34{a_1} + 68d = 18{a_1} + 72d
⇒16a1=4d\Rightarrow 16{a_1} = 4d

∴\therefore

d=4a1d = 4{a_1}

And

110110

\therefore

110 ∴\therefore

a1=2{a_1} = 2

(∵\because

ai ∈N{a_i}\, \in N

)

d=8d = 8

∴\therefore

S10=5[4+9×8]=380{S_{10}} = 5[4 + 9 \times 8] = 380
Q49
Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5 . Let the sum of its first five terms be 9825\dfrac{98}{25}. Then the sum of the first 21 terms of an AP, whose first term is 10ar,nth 10\mathrm{a r}, \mathrm{n}^{\text{th }} term is an\mathrm{a}_{\mathrm{n}} and the common difference is 10ar210 \mathrm{ar}^{2}, is equal to :
A 21 a1121 \,\mathrm{a}_{11}
B 22 a1122 \,\mathrm{a}_{11}
C 15 a1615 \,\mathrm{a}_{16}
D 14 a1614 \,\mathrm{a}_{16}
Correct Answer
Option A
Solution

Let first term of G.P. be a and common ratio is r Then,

a1−r=5{a \over {1 - r}} = 5

...... (i)

a(r5−1)(r−1)=9825⇒1−r5=98125a{{({r^5} - 1)} \over {(r - 1)}} = {{98} \over {25}} \Rightarrow 1 - {r^5} = {{98} \over {125}}

∴\therefore

r5=27125, r=(35)35{r^5} = {{27} \over {125}},\,r = {\left( {{3 \over 5}} \right)^{{3 \over 5}}}

∴\therefore Then,

S21=212[2×10ar+20×10ar2]{S_{21}} = {{21} \over 2}\left[ {2 \times 10ar + 20 \times 10a{r^2}} \right]
=21[10ar+10 . 10ar2]= 21\left[ {10ar + 10\,.\,10a{r^2}} \right]
=21 a11= 21\,{a_{11}}
Q50
Let x, y > 0. If x3y2 = 215, then the least value of 3x + 2y is
A 30
B 32
C 36
D 40
Correct Answer
Option D
Solution

x, y > 0 and x3y2 = 215 Now, 3x + 2y = (x + x + x) + (y + y) So, by A.M ≥\ge G.M inequality

3x+2y5≥x3 . y25{{3x + 2y} \over 5} \ge \sqrt[5]{{x^3}\,.\,{y^2}}

∴\therefore

3x+2y≥52155≥403x + 2y \ge 5\sqrt[5]{{2^{15}}} \ge 40

∴\therefore Least value of

3x+4y=403x + 4y = 40
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