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Sequences and Series

JEE Mathematics · 201 questions · Page 12 of 21 · Click an option or "Show Solution" to reveal answer

Q111
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 :
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.
Correct Answer
Option B
Solution

(a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0 ⇒\Rightarrow (a2p2 + 2abp + b2 ) + (b2p2 + 2bcp + c2 ) + (c2 p2 + 2cdp + d2) = 0 ⇒\Rightarrow (ab + b)2 + (bp + c)2 + (cp + d)2 = 0 Note : If sum of two or more positive quantity is zero then they are all zero. ∴\therefore ap + b = 0 and bp + c = 0 and cp + d = 0 p =

−ba- {b \over a}

=

−cb- {c \over b}

=

−dc- {d \over c}

or

ba{b \over a}

=

cb{c \over b}

=

dc{d \over c}

∴\therefore a, b, c, d are in G.P.

Q112
The sum of the series 1x+1+2x2+1+22x4+1+......+2100x2100+1{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+21014101−11 + {{{2^{101}}} \over {{4^{101}} - 1}}
B 1+21004101−11 + {{{2^{100}}} \over {{4^{101}} - 1}}
C 1−21004100−11 - {{{2^{100}}} \over {{4^{100}} - 1}}
D 1−21012400−11 - {{{2^{101}}} \over {{2^{400}} - 1}}
Correct Answer
Option D
Solution
S=1x+1+2x2+1+22x4+1+......+2100x2100+1S = {1 \over {x + 1}} + {2 \over {{x^2} + 1}} + {{{2^2}} \over {{x^4} + 1}} + ...... + {{{2^{100}}} \over {{x^{{2^{100}}}} + 1}}
S+11−x=11−x+1x+1+......=21−x2+21+x2+....S + {1 \over {1 - x}} = {1 \over {1 - x}} + {1 \over {x + 1}} + ...... = {2 \over {1 - {x^2}}} + {2 \over {1 + {x^2}}} + ....
S+11−x=21011−x400S + {1 \over {1 - x}} = {{{2^{101}}} \over {1 - {x^{400}}}}
S=1−21012400−1S = 1 - {{{2^{101}}} \over {{2^{400}} - 1}}
Q113
The sum of the infinite series 1+56+1262+2263+3564+5165+7066+  .....1 + {5 \over 6} + {{12} \over {{6^2}}} + {{22} \over {{6^3}}} + {{35} \over {{6^4}}} + {{51} \over {{6^5}}} + {{70} \over {{6^6}}} + \,\,..... is equal to :
A 425216{{425} \over {216}}
B 429216{{429} \over {216}}
C 288125{{288} \over {125}}
D 280125{{280} \over {125}}
Correct Answer
Option C
Solution
S=1+56+1262+2263+S = 1 + {5 \over 6} + {{12} \over {{6^2}}} + {{22} \over {{6^3}}} +

...... (1)

16S=16+562+1263+{1 \over 6}S = {1 \over 6} + {5 \over {{6^2}}} + {{12} \over {{6^3}}} +

...... (2)

S−16S=1+46+762+1063+S - {1 \over 6}S = 1 + {4 \over 6} + {7 \over {{6^2}}} + {{10} \over {{6^3}}} +

........

⇒5S6=1+46+762+1063+\Rightarrow {{5S} \over 6} = 1 + {4 \over 6} + {7 \over {{6^2}}} + {{10} \over {{6^3}}} +

....... (3) Now, multiplying both sides by

16{1 \over 6}

, we get

⇒5S36=16+462+763+1064+\Rightarrow {{5S} \over {36}} = {1 \over 6} + {4 \over {{6^2}}} + {7 \over {{6^3}}} + {{10} \over {{6^4}}} +

...... (4) Subtract equation (4) from equation (3), we get

2536S=1+36+362+363+{{25} \over {36}}S = 1 + {3 \over 6} + {3 \over {{6^2}}} + {3 \over {{6^3}}} +

......

⇒25S36=1+351−16\Rightarrow {{25S} \over {36}} = 1 + {{{3 \over 5}} \over {1 - {1 \over 6}}}
=1+36×65= 1 + {3 \over 6} \times {6 \over 5}
=1+35=85= 1 + {3 \over 5} = {8 \over 5}
⇒S=85×3625=288125\Rightarrow S = {8 \over 5} \times {{36} \over {25}} = {{288} \over {125}}
Q114
Let Sn = 1 . (n −- 1) + 2 . (n −- 2) + 3 . (n −- 3) + ..... + (n −- 1) . 1, n ≥\ge 4. The sum ∑n=4∞(2Snn!−1(n−2)!)\sum\limits_{n = 4}^\infty {\left( {{{2{S_n}} \over {n!}} - {1 \over {(n - 2)!}}} \right)} is equal to :
A e−13{{e - 1} \over 3}
B e−26{{e - 2} \over 6}
C e3{e \over 3}
D e6{e \over 6}
Correct Answer
Option A
Solution

Let Tr = r(n −- r) Tr = nr −- r2

⇒Sn=∑r=1nTr=∑r=1n(nr−r2)\Rightarrow {S_n} = \sum\limits_{r = 1}^n {{T_r} = \sum\limits_{r = 1}^n {(nr - {r^2})} }
Sn=n . (n)(n+1)2−n(n+1)(2n+1)6{S_n} = {{n\,.\,(n)(n + 1)} \over 2} - {{n(n + 1)(2n + 1)} \over 6}
Sn=n(n−1)(n+1)6{S_n} = {{n(n - 1)(n + 1)} \over 6}

Now,

∑n=4∞(2Snn!−1(n−2)!)\sum\limits_{n = 4}^\infty {\left( {{{2{S_n}} \over {n!}} - {1 \over {(n - 2)!}}} \right)}
=∑r=4∞(2.n(n−1)(n+1)6 . n(n−1)(n−2)!−1(n−2)!)= \sum\limits_{r = 4}^\infty {\left( {2.{{n(n - 1)(n + 1)} \over {6\,.\,n(n - 1)(n - 2)!}} - {1 \over {(n - 2)!}}} \right)}
=∑r=4∞(13(n−2+3(n−2)!)−1(n−2)!)= \sum\limits_{r = 4}^\infty {\left( {{1 \over 3}\left( {{{n - 2 + 3} \over {(n - 2)!}}} \right) - {1 \over {(n - 2)!}}} \right)}
=∑r=4∞13.1(n−3)!=13(e−1)= \sum\limits_{r = 4}^\infty {{1 \over 3}.{1 \over {(n - 3)!}} = {1 \over 3}(e - 1)}

Option (a)

Q115
The sum ∑k=120k12k\sum\limits_{k = 1}^{20} {k{1 \over {{2^k}}}} is equal to
A 2−112192 - {11 \over {{2^{19}}}}
B 2−32172 - {3 \over {{2^{17}}}}
C 1−112201 - {11 \over {{2^{20}}}}
D 2−212202 - {21 \over {{2^{20}}}}
Correct Answer
Option A
Solution

Let S =

∑k=120k12k\sum\limits_{k = 1}^{20} {k{1 \over {{2^k}}}}

⇒\Rightarrow S =

12+222+323+424+.......+20220{1 \over 2} + {2 \over {{2^2}}} + {3 \over {{2^3}}} + {4 \over {{2^4}}} + ....... + {{20} \over {{2^{20}}}}

.......(

1) This is an Arithmetic Geometric Sequence.

Here numerator is in A.P and denominator is in G.P.

To solve Arithmetic Geometric Sequence, we multiply the series S with the common ratio of G.P.

Here common ratio of G.P is

12{1 \over 2}

. By multiplying (1) with

12{1 \over 2}

we get,

S2=122+223+324+425+.......+19220+20221{S \over 2} = {1 \over {{2^2}}} + {2 \over {{2^3}}} + {3 \over {{2^4}}} + {4 \over {{2^5}}} + ....... + {{19} \over {{2^{20}}}} + {{20} \over {{2^{21}}}}

....(2) Subtract (2) from (1), S -

S2{S \over 2}

=

12+122+123+124+.......+1220−20221{1 \over 2} + {1 \over {{2^2}}} + {1 \over {{2^3}}} + {1 \over {{2^4}}} + ....... + {1 \over {{2^{20}}}} - {{20} \over {{2^{21}}}}

⇒\Rightarrow

S2{S \over 2}

=

12(1−1220)1−12{{{1 \over 2}\left( {1 - {1 \over {{2^{20}}}}} \right)} \over {1 - {1 \over 2}}}

-

20221{{20} \over {{2^{21}}}}

⇒\Rightarrow S =

2(1−1220)−202202\left( {1 - {1 \over {{2^{20}}}}} \right) - {{20} \over {{2^{20}}}}

⇒\Rightarrow S =

2−2220−202202 - {2 \over {{2^{20}}}} - {{20} \over {{2^{20}}}}

⇒\Rightarrow S =

2−222202 - {{22} \over {{2^{20}}}}

⇒\Rightarrow S =

2−112192 - {{11} \over {{2^{19}}}}
Q116
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 52{5 \over 2}
B 12{1 \over 2}
C 252{25 \over 2}
D 92{9 \over 2}
Correct Answer
Option B
Solution

Sum of infinite terms :

a1−r=15{a \over {1 - r}} = 15

..... (i) Series formed by square of terms : a2, a2r2, a2r4, a2r6 ....... Sum =

a21−r2=150{{{a^2}} \over {1 - {r^2}}} = 150
⇒a1−r.a1+r=150⇒15.a1+r=150\Rightarrow {a \over {1 - r}}.{a \over {1 + r}} = 150 \Rightarrow 15.{a \over {1 + r}} = 150
⇒a1+r=10\Rightarrow {a \over {1 + r}} = 10

...... (ii) by (i) and (ii), a = 12; r =

15{1 \over 5}

Now, series : ar2, ar4, ar6 Sum =

ar21−r2=12.(125)1−125=12{{a{r^2}} \over {1 - {r^2}}} = {{12.\left( {{1 \over {25}}} \right)} \over {1 - {1 \over {25}}}} = {1 \over 2}
Q117
Let the first term α\alpha and the common ratio r of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to
A 241
B 231
C 220
D 210
Correct Answer
Option B
Solution

Given that the first term aa and common ratio rr of a geometric progression be positive integer.

So, their 1st three terms are a,ar,ar2a, a r, a r^2 According to the question, a2+a2r2+a2r4=33033a^2+a^2 r^2+a^2 r^4=33033

⇒a2(1+r2+r4)=3×7×11×11×13=3×7×13×112\begin{aligned} \Rightarrow a^2\left(1+r^2+r^4\right) & =3 \times 7 \times 11 \times 11 \times 13 \\ & =3 \times 7 \times 13 \times 11^2 \end{aligned}
∴a2=112⇒a=11 and 1+r2+r4=273⇒r2+r4=272⇒r4+r2−272=0⇒(r2+17)(r2−16)=0⇒r2=−17(not possible),⇒r2−16=0 ⇒r=±4⇒r=4  (∵r>0)\begin{aligned} & \therefore \quad a^2=11^2 \\\\ & \Rightarrow \quad a=11 \\\\ & \text{ and } 1+r^2+r^4=273 \\\\ & \Rightarrow r^2+r^4=272 \\\\ & \Rightarrow r^4+r^2-272=0 \\\\ & \Rightarrow\left(r^2+17\right)\left(r^2-16\right)=0 \\\\ & \Rightarrow r^2=-17(not ~possible),\\\\ & \Rightarrow r^2-16=0 \\\\ & \text{ } \Rightarrow r= \pm 4 \\\\ & \Rightarrow r=4 ~~( \because r > 0) \end{aligned}

So, sum of these first three terms is a+ar+ar2a+a r+a r^2 =11+44+176=231=11+44+176=231

Q118
The sum of the first 2020 terms of the series 5+11+19+29+41+…5+11+19+29+41+\ldots is :
A 3420
B 3450
C 3250
D 3520
Correct Answer
Option D
Solution
Sn=5+11+19+29+41+….+TnSn=        5+11+19+29+….+Tn−1+Tn0=5+6+8+10+12+…..Tn\begin{aligned} & \mathrm{S}_n=5+11+19+29+41+\ldots .+\mathrm{T}_n \\\\ & \mathrm{S}_n=~~~~~~~~ 5+11+19+29+\ldots .+\mathrm{T}_{n-1}+\mathrm{T}_n \\\\ & \\ 0=5+6+8+10+12+\ldots . . \mathrm{T}_n \end{aligned}
0=5+n−12[2×6+(n−2)(2)]−Tn⇒Tn=5+(n−1)(n+4)⇒Tn=5+n2+3n−4⇒Tn=n2+3n+1ΣTn=Σn2+3Σn+Σ1\begin{array}{rlrl} &0 =5+\frac{n-1}{2}[2 \times 6+(n-2)(2)]-T_n \\\\ &\Rightarrow T_n =5+(n-1)(n+4) \\\\ &\Rightarrow T_n =5+n^2+3 n-4 \\\\ &\Rightarrow T_n =n^2+3 n+1 \\\\ &\Sigma T_n =\Sigma n^2+3 \Sigma n+\Sigma 1 \end{array}
⇒Sn=n(n+1)(2n+1)6+3n(n+1)2+n\Rightarrow \quad S_n=\frac{n(n+1)(2 n+1)}{6}+\frac{3 n(n+1)}{2}+n

When, n=20n=20 Then,

S20=20×21×416+3×20×212+20=2870+630+20=3520\begin{aligned} S_{20} & =\frac{20 \times 21 \times 41}{6}+\frac{3 \times 20 \times 21}{2}+20 \\\\ & =2870+630+20=3520 \end{aligned}
Q119
If 11+2+12+3+…+199+100=m\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\dfrac{1}{\sqrt{99}+\sqrt{100}}=m and 11⋅2+12⋅3+…+199⋅100=n\dfrac{1}{1 \cdot 2}+\dfrac{1}{2 \cdot 3}+\ldots+\dfrac{1}{99 \cdot 100}=\mathrm{n}, then the point (m,n)(\mathrm{m}, \mathrm{n}) lies on the line
A 11(x−1)−100y=011(x-1)-100 y=0
B 11x−100y=011 x-100 y=0
C 11(x−1)−100(y−2)=011(x-1)-100(y-2)=0
D 11(x−2)−100(y−1)=011(x-2)-100(y-1)=0
Correct Answer
Option B
Solution
11+2+12+3+…+199+100=m and 11⋅2+12⋅3+…+199⋅100=n11+2+12+3+…+199+100=11+2×2−12−1+13+2×3−23−2+…+199+100×100−99100−99\begin{aligned} & \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m \\ & \text{ and } \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n \\ & \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}} \\ & =\frac{1}{\sqrt{1}+\sqrt{2}} \times \frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ & \quad+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}} \times \frac{\sqrt{100}-\sqrt{99}}{\sqrt{100}-\sqrt{99}} \end{aligned}
=2−1+3−2+…+100−99=100−1=10−1⇒m=9\begin{aligned} & =\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{100}-\sqrt{99} \\ & =\sqrt{100}-\sqrt{1} \\ & =10-1 \\ & \Rightarrow m=9 \end{aligned}

and

11⋅2+12⋅3+…+199⋅100=n\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n
2−11×2+3−22×3+…+100−99100×99=n⇒1−12+12−13+…+199−1100=n⇒n=1−1100⇒n=99100(m,n)=(9,99100)\begin{aligned} & \frac{2-1}{1 \times 2}+\frac{3-2}{2 \times 3}+\ldots+\frac{100-99}{100 \times 99}=n \\ & \Rightarrow 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\ldots+\frac{1}{99}-\frac{1}{100}=n \\ & \Rightarrow n=1-\frac{1}{100} \\ & \Rightarrow n=\frac{99}{100} \\ & (m, n)=\left(9, \frac{99}{100}\right) \end{aligned}

Satisfies the line

11x−100y=011 x-100 y=0
Q120
If an=−24n2−16n+15{a_n} = {{ - 2} \over {4{n^2} - 16n + 15}}, then a1+a2 + .... + a25{a_1} + {a_2}\, + \,....\, + \,{a_{25}} is equal to :
A 51144{{51} \over {144}}
B 49138{{49} \over {138}}
C 50141{{50} \over {141}}
D 52147{{52} \over {147}}
Correct Answer
Option C
Solution
∑i=125ai=∑−24n2−16n+15=∑−2(2n−5)(2n−3)\sum\limits_{i = 1}^{25} {{a_i} = \sum {{{ - 2} \over {4{n^2} - 16n + 15}} = \sum {{{ - 2} \over {(2n - 5)(2n - 3)}}} } }
=∑i=125(12n−3−12n−5)= \sum\limits_{i = 1}^{25} {\left( {{1 \over {2n - 3}} - {1 \over {2n - 5}}} \right)}
=[(1−1−1−3)+(11−1−1)+(13−11)......= \left[ {\left( {{1 \over { - 1}} - {1 \over { - 3}}} \right) + \left( {{1 \over 1} - {1 \over { - 1}}} \right) + \left( {{1 \over 3} - {1 \over 1}} \right)......} \right.
=12(25)−3+13=50141= {1 \over {2(25) - 3}} + {1 \over 3} = {{50} \over {141}}
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