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Sequences and Series

JEE Mathematics · 201 questions · Page 2 of 21 · Click an option or "Show Solution" to reveal answer

Q11
Let a1{a_1}, a2{a_2}, a3{a_3}.....be terms on A.P. If a1+a2+.....apa1+a2+.....aq=p2q2, p≠q, then a6a21 {{{a_1} + {a_2} + .....{a_p}} \over {{a_1} + {a_2} + .....{a_q}}} = {{{p^2}} \over {{q^2}}},\,p \ne q,\,then\,{{{a_6}} \over {{a_{21}}}}\, equals
A 4111{{41} \over {11}}
B 72{7 \over 2}
C 27{2 \over 7}
D 1141{{11} \over {41}}
Correct Answer
Option D
Solution
p2[2a1+(p−1)d]q2[2a1+(q−1)d]=p2q2{{{p \over 2}\left[ {2{a_1} + \left( {p - 1} \right)d} \right]} \over {{q \over 2}\left[ {2{a_1} + \left( {q - 1} \right)d} \right]}} = {{{p^2}} \over {{q^2}}}
⇒2a1+(p−1)d2a1+(p−1)d=pq\Rightarrow {{2{a_1} + \left( {p - 1} \right)d} \over {2{a_1} + \left( {p - 1} \right)d}} = {p \over q}
a1+(p−12)da1+(q−12)d=pq{{{a_1} + \left( {{{p - 1} \over 2}} \right)d} \over {{a_1} + \left( {{{q - 1} \over 2}} \right)d}} = {p \over q}

For

a6a21,  p=11, q=41{{{a_6}} \over {a{}_{21}}},\,\,p = 11,\,q = 41
⇒a6a21=1141\Rightarrow {{{a_6}} \over {a{}_{21}}} = {{11} \over {41}}
Q12
The sum to infinite term of the series 1+23+632+1033+1434+.....1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + {{14} \over {{3^4}}} + ..... is
A 3
B 4
C 6
D 2
Correct Answer
Option A
Solution

We have

S=1+23+632+1033+1434+........∞     ...(1)S = 1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + {{14} \over {{3^4}}} + ........\infty \,\,\,\,\,...\left( 1 \right)

Multiplying both sides by

13{1 \over 3}

we get

13S=13+232+633+1034+.......     ...(2){1 \over 3}S = {1 \over 3} + {2 \over {{3^2}}} + {6 \over {{3^3}}} + {{10} \over {{3^4}}} + .......\,\,\,\,\,...\left( 2 \right)

Subtracting eqn.

(2)(2)

from eqn.

(1)(1)

we get

23S=1+13+432+433+434+.....∞{2 \over 3}S = 1 + {1 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + {4 \over {{3^4}}} + .....\infty
⇒23S=43+432+433+434+.....∞\Rightarrow {2 \over 3}S = {4 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + {4 \over {{3^4}}} + .....\infty
⇒23S=431−13=43×32\Rightarrow {2 \over 3}S = {{{4 \over 3}} \over {1 - {1 \over 3}}} = {4 \over 3} \times {3 \over 2}
⇒S−3\Rightarrow S - 3
Q13
A person is to count 4500 currency notes. Let an{a_n} denote the number of notes he counts in the nth{n^{th}} minute. If a1{a_1} = a2{a_2} = ....= a10{a_{10}}= 150 and a10{a_{10}}, a11{a_{11}},.... are in an AP with common difference - 2, then the time taken by him to count all notes is
A 34 minutes
B 125 minutes
C 135 minutes
D 24 minutes
Correct Answer
Option A
Solution

Till

1010

th minute number of counted notes

=1500= 1500
3000=n2[2×148+(n−1)(−2)]3000 = {n \over 2}\left[ {2 \times 148 + \left( {n - 1} \right)\left( { - 2} \right)} \right]
                \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
=n[148−n+1]= n\left[ {148 - n + 1} \right]

⇒\Rightarrow

n2−149n+3000=0{n^2} - 149n + 3000 = 0
⇒n=125,24\Rightarrow n = 125,24

But

n=125n=125

is not possible ∴\therefore total time

=24+10=34= 24 + 10 = 34

minutes.

Q14
Statement-1: The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) +.....+ (361 + 380 + 400) is 8000. Statement-2: ∑k=1n(k3−(k−1)3)=n3\sum\limits_{k = 1}^n {\left( {{k^3} - {{(k - 1)}^3}} \right)} = {n^3}, for any natural number n.
A Statement-1 is false, Statement-2 is true.
B Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
C Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
D Statement-1 is true, Statement-2 is false.
Correct Answer
Option B
Solution
nn

th term of the given series

=Tn=(n−1)2+(n−1)n+n2= {T_n} = {\left( {n - 1} \right)^2} + \left( {n - 1} \right)n + {n^2}
=((n−1)3−n3)(n−1)−n= {{\left( {{{\left( {n - 1} \right)}^3} - {n^3}} \right)} \over {\left( {n - 1} \right) - n}}
=n3−(n−1)3= {n^3} - {\left( {n - 1} \right)^3}
⇒Sn=∑k=1n[k3−(k−1)3]\Rightarrow {S_n} = \sum\limits_{k = 1}^n {\left[ {{k^3} - {{\left( {k - 1} \right)}^3}} \right]}
⇒8000=n3\Rightarrow 8000 = {n^3}
⇒n=20  \Rightarrow n = 20\,\,

which is a natural number. Now, put

n=1,2,3,.....20n = 1,2,3,.....20
T1=13−03{T_1} = {1^3} - {0^3}
T2=23−13{T_2} = {2^3} - {1^3}

. . .

T20=203−193{T_{20}} = {20^3} - {19^3}

Now,

T1+T2+.....+T20=S20{T_1} + {T_2} + ..... + {T_{20}} = {S_{20}}
⇒S20=203−03=8000\Rightarrow {S_{20}} = {20^3} - {0^3} = 8000

Hence, both the given statements are true and statement

22

supports statement

1.1.
Q15
Let the sum of the first n terms of a non-constant A.P., a1, a2, a3, ..... be 50n+n(n−7)2A50n + {{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)
Correct Answer
Option C
Solution

Sn =

50n+n(n−7)2A50n + {{n(n - 7)} \over 2}A

We know, nth tem Tn = Sn - Sn - 1 =

50n+n(n−7)2A50n + {{n(n - 7)} \over 2}A

-

50(n−1)−(n−1)(n−8)2A50\left( {n - 1} \right) - {{\left( {n - 1} \right)\left( {n - 8} \right)} \over 2}A

= 50 +

A2[n2−7n−n2+9n−8]{A \over 2}\left[ {{n^2} - 7n - {n^2} + 9n - 8} \right]

= 50 + A(n - 4) We also know, common difference d = Tn - Tn - 1 = 50 + A(n - 4) - 50 - A(n - 5) = A And T50 = 50 + A(50 - 4) = 50 + 46A ∴\therefore (d, a50) = (A, 50+46A)

Q16
The sum of first 9 terms of the series. 131+13+231+3+13+23+331+3+5+......{{{1^3}} \over 1} + {{{1^3} + {2^3}} \over {1 + 3}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 3 + 5}} + ......
A 142
B 192
C 71
D 96
Correct Answer
Option D
Solution
nth{n^{th}}

term of series

=[n(n+1)2]n2=14(n+1)2= {{\left[ {{{n\left( {n + 1} \right)} \over 2}} \right]} \over {{n^2}}} = {1 \over 4}{\left( {n + 1} \right)^2}

Sum of

nn

term

=∑14(n+1)2= \sum {{1 \over 4}} {\left( {n + 1} \right)^2}
=14[∑n2+2∑n+n]= {1 \over 4}\left[ {\sum {n{}^2} + 2\sum n + n} \right]
=14[n(n+1)(2n+1)6+2n(n+1)2+n]= {1 \over 4}\left[ {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6} + {{2n\left( {n + 1} \right)} \over 2} + n} \right]

Sum of

99

terms

=14[9×10×196+18×102+9]=3844=96= {1 \over 4}\left[ {{{9 \times 10 \times 19} \over 6} + {{18 \times 10} \over 2} + 9} \right] = {{384} \over 4} = 96
Q17
For any three positive real numbers a, b and c, 9(25a2{a^2} + b2) + 25(c2 - 3aac) = 15b(3aa + c). Then
A b, c and aa are in G.P.
B b, c and aa are in A.P.
C aa, b and c are in A.P.
D aa, b and c are in G.P.
Correct Answer
Option B
Solution

9(25

a2{a^2}

+ b2) + 25(c2 - 3

aa

c) = 15b(3

aa

+ c)

⇒225a2+9b2+25c2−75ac=45ab+15bc\Rightarrow 225{a^2} + 9{b^2} + 25{c^2} - 75ac = 45ab + 15bc
⇒(15a)2+(3b)2+(5c)2−75ac=45ab+15bc\Rightarrow {\left( {15a} \right)^2} + {\left( {3b} \right)^2} + {\left( {5c} \right)^2} - 75ac = 45ab + 15bc

⇒\Rightarrow

12[(15a−3b)2+(3b−5c)2+(5c−15a)2]=0{1 \over 2}\left[ {{{\left( {15a - 3b} \right)}^2} + {{\left( {3b - 5c} \right)}^2} + {{\left( {5c - 15a} \right)}^2}} \right] = 0

it is possible when 15a – 3b = 0, 3b – 5 c = 0 and 5c – 15a = 0 ⇒\Rightarrow 15a = 3b = 5c ⇒\Rightarrow b =

5c3{{5c} \over 3}

, a =

c3{c \over 3}

⇒\Rightarrow a + b =

c3+5c3{c \over 3} + {{5c} \over 3}

=

6c3{{6c} \over 3}

= 2c ∴\therefore b, c, a are in A.P.

Q18
Let a1{a_1}, a2{a_2}, a3{a_3}, ......... ,a49{a_{49}} be in A.P. such that ∑k=012a4k+1=416\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416 and a9+a43=66{a_9} + {a_{43}} = 66. a12+a22+.......+a172=140ma_1^2 + a_2^2 + ....... + a_{17}^2 = 140m, then m is equal to
A 33
B 66
C 68
D 34
Correct Answer
Option D
Solution

a1, a2, a3 . . . a43 are in AP So, a2 = a1 + d a3 = a1 + 2d . . . a49 =a1 + 48d Now given,

a9+a43=66{a_9} + {a_{43}} = 66
⇒    \Rightarrow \,\,\,\,

a1 + 8d + a1 + 42d = 66

⇒    \Rightarrow \,\,\,\,

2a1 + 50d = 66

⇒    \Rightarrow \,\,\,\,

a1 + 25d = 33 . . . . . (1)

∑k=012a4k+1\sum\limits_{k = 0}^{12} {{a_{4k + 1}}}

= 416

⇒    \Rightarrow \,\,\,\,

a1 + a5 + a9 + a13 +. . . . . 13 items = 416

⇒    \Rightarrow \,\,\,\,

a1 + a1 + 4d + a1 + 8d + . . . . a1 + 48d = 416

⇒    \Rightarrow \,\,\,\,

13a1 + 4d +8d + 12d + . . . . . 48d = 416

⇒    \Rightarrow \,\,\,\,

13a1 + 4 (1+ 2 + 3 + . . . + 12) d = 416

⇒    13  a1+ 4  × 12×132×\Rightarrow \,\,\,\,13\,\,a{}_1 + \,4\,\, \times \,{{12 \times 13} \over 2} \times

d = 416

⇒    \Rightarrow \,\,\,\,

13a1 + 24 ×\times 13d = 416

⇒    \Rightarrow \,\,\,\,

a1 + 24 d =32 . . . .(2) Solving (1) and (2) we get, d = 1 and

a1=8{a_1} = 8
∴   \therefore\,\,\,

a1 = 8 a2 = 8 + 1 = 9 a3 = 8 + 2 = 10 . . . a17 = 8 + 16 = 24 Now,

a12+a22+......  +a172  =  140ma_1^2 + a{}_2^2 + ......\,\, + a_{17}^2\,\, = \,\,140m
⇒    \Rightarrow \,\,\,\,
a12+a22+......  +a172=140 ma_1^2 + a{}_2^2 + ......\,\, + a_{17}^2 = 140\,m
⇒     82 +  92 + 102+......(24)2=140 m\Rightarrow \,\,\,\,\,{8^2}\, + \,\,{9^2}\, + \,{10^2} + ......{(24)^2} = 140\,m

We can write above series like this,

⇒     \Rightarrow \,\,\,\,\,

(12 +22 + . . . . +242) −- (12 + 22 + . . . . .+ 72) = 140 m

⇒24(25)(49)6−7×8×156=140 m\Rightarrow {{24\left( {25} \right)\left( {49} \right)} \over 6} - {{7 \times 8 \times 15} \over 6} = 140\,m
⇒     \Rightarrow \,\,\,\,\,

490 −- 140 = 140 m

⇒     \Rightarrow \,\,\,\,\,

4760 = 140 m

⇒     \Rightarrow \,\,\,\,\,

m = 34

Q19
Let aa, b and c be in G.P. with common ratio r, where aa ≠ \ne 0 and 0 < r ≤ \le 12{1 \over 2} . If 3aa, 7b and 15c are the first three terms of an A.P., then the 4th term of this A.P. is :
A aa
B 73a{7 \over 3}a
C 5aa
D 23a{2 \over 3}a
Correct Answer
Option A
Solution

a = a, b = ar and c = ar2 3a, 7b, 15c →\to A.P. 14b = 3a + 15c 14(ar) = 3a + 15(ar2) 15r2 – 14r + 3 = 0

⇒r=13,35(rejected)\Rightarrow r = {1 \over 3},{3 \over 5}(rejected)

Common difference = 7b – 3a = 7ar – 3a ⇒\Rightarrow

7a3−3a=−23a{{7a} \over 3} - 3a = - {2 \over 3}a

4th term is ⇒\Rightarrow

15c−23a=159a−23a=a15c - {2 \over 3}a = {{15} \over 9}a - {2 \over 3}a = a
Q20
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
Correct Answer
Option B
Solution

a1 + a7 + a16 = 40

a1+(a1+6d)+(a1+15d)=40{a_1} + \left( {{a_1} + 6d} \right) + ({a_1} + 15d) = 40
⇒3a1+21d=40\Rightarrow 3{a_1} + 21d = 40
⇒a1+7d=403\Rightarrow {a_1} + 7d = {{40} \over 3}
⇒a1+a2.......+a15=152[a1+a15]\Rightarrow {a_1} + {a_2}....... + {a_{15}} = {{15} \over 2}[{a_1} + {a_{15}}]
⇒152[a1+a1+14d]⇒15(a1+7d)=15×403=200\Rightarrow {{15} \over 2}[{a_1} + {a_1} + 14d] \Rightarrow 15({a_1} + 7d) = 15 \times {{40} \over 3} = 200
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