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Sequences and Series

JEE Mathematics · 201 questions · Page 11 of 21 · Click an option or "Show Solution" to reveal answer

Q101
If a1,a2,....an{{a_1},{a_2},....{a_n}} are in H.P., then the expression a1 a2+ a2 a3 +....+an−1 an{{a_1}\,{a_2} + \,{a_2}\,{a_3}\, + .... + {a_{n - 1}}\,{a_n}} is equal to
A n(a1 −an)n({a_1}\, - {a_n})
B (n−1)(a1 −an)(n - 1)({a_1}\, - {a_n})
C na1ann{a_1}{a_n}
D (n−1)  a1an(n - 1)\,\,{a_1}{a_n}
Correct Answer
Option D
Solution
1a2−1a1=1a3−1a2=.........{1 \over {{a_2}}} - {1 \over {{a_1}}} = {1 \over {{a_3}}} - {1 \over {{a_2}}} = .........
               =1an−1an−1=d\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over {{a_n}}} - {1 \over {{a_{n - 1}}}} = d

(say) Then

a1a2=a1−a2d, a2a3=a2−a3d,{a_1}{a_2} = {{{a_1} - a{}_2} \over d},\,{a_2}{a_3} = {{{a_2} - {a_3}} \over d},
               ......, an−1an=an−1−and\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......,\,{a_{n - 1}}{a_n} = {{{a_{n - 1}} - {a_n}} \over d}

∴\therefore

a1a2+a2a3+.........+an−1an{a_1}a{}_2 + {a_2}{a_3} + ......... + {a_{n - 1}}{a_n}
=a1−a2d+a2−a3d+......= {{{a_1} - {a_2}} \over d} + {{{a_2} - {a_3}} \over d} + ......
               +an−1−and\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {{{a_{n - 1}} - {a_n}} \over d}
=1a[a1−a2+a2−a3+.......= {1 \over a}\left[ {{a_1}} \right. - {a_2} + {a_2} - {a_3} + .......
               +an−1−an]=a1−and\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. { + {a_{n - 1}} - an} \right] = {{{a_1} - {a_n}} \over d}

Also,

1an=1a1+(n−1)d{1 \over {{a_n}}} = {1 \over {{a_1}}} + \left( {n - 1} \right)d
⇒a1−ana1an=(n−1)d\Rightarrow {{{a_1} - {a_n}} \over {{a_1}{a_n}}} = \left( {n - 1} \right)d
⇒a1−and=(n−1)a1an\Rightarrow {{{a_1} - {a_n}} \over d} = \left( {n - 1} \right){a_1}{a_n}

Which is the required result.

Q102
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:
A (2480, 249)
B (2480, 248)
C (2490, 248)
D (2490, 249)
Correct Answer
Option C
Solution
an=a1+(n−1)d{a_n} = {a_1} + (n - 1)d
⇒300=1+(n−1)d\Rightarrow 300 = 1 + (n - 1)d
⇒(n−1)d=299=13×23\Rightarrow (n - 1)d = 299 = 13 \times 23

since, n

∈\in

[15, 50] ∴\therefore n = 24 and d = 13

an−4=a20=1+19×13=248{a_{n - 4}} = {a_{20}} = 1 + 19 \times 13 = 248
⇒an−4=248\Rightarrow {a_{n - 4}} = 248
Sn−4=202{1+248}=2490{S_{n - 4}} = {{20} \over 2}\{ 1 + 248\} = 2490
Q103
Let a1, a2, a3, ..... a10 be in G.P. with ai > 0 for i = 1, 2, ….., 10 and S be the set of pairs (r, k), r, k ∈ \in N (the set of natural numbers) for which ∣log⁡e a1ra2klog⁡e a2ra3klog⁡e a3ra4klog⁡e a4ra5klog⁡e a5ra6klog⁡e a6ra7klog⁡e a7ra8klog⁡e a8ra9klog⁡e a9ra10k∣\left| \begin{array}{lll}{{{\log }_e}\,{a_1}^r{a_2}^k} & {{{\log }_e}\,{a_2}^r{a_3}^k} & {{{\log }_e}\,{a_3}^r{a_4}^k} \\ {{{\log }_e}\,{a_4}^r{a_5}^k} & {{{\log }_e}\,{a_5}^r{a_6}^k} & {{{\log }_e}\,{a_6}^r{a_7}^k} \\ {{{\log }_e}\,{a_7}^r{a_8}^k} & {{{\log }_e}\,{a_8}^r{a_9}^k} & {{{\log }_e}\,{a_9}^r{a_{10}}^k} \end{array} \right| == 0. Then the number of elements in S, is -
A 10
B 4
C 2
D infinitely many
Correct Answer
Option D
Solution

Apply C3 →\to C3 −- C2 C2 →\to C2 −- C1 We get D = 0

Q104
If 210 + 29.31 + 28 .32 +.....+ 2.39 + 310 = S - 211, then S is equal to :
A 3112+210{{{3^{11}}} \over 2} + {2^{10}}
B 311 — 212
C 2.311
D 311
Correct Answer
Option D
Solution

Let S1 = 210 + 29.31 + 28 .32 +.....+ 2.39 + 310 ....(1) Also

3S12{{3{S_1}} \over 2}

= 29.31 + 28 .32 +.....+ 2.39 + 310 +

3112{{{3^{11}}} \over 2}

......(2) Performing (1) - (2), we get

−S12=210−3112- {{{S_1}} \over 2} = {2^{10}} - {{{3^{11}}} \over 2}

⇒\Rightarrow S1 = 311 - 211 According to question, 311 - 211 = S - 211 ⇒\Rightarrow S = 311

Q105
A man saves ₹ 200 in each of the first three months of his service. In each of the subsequent months his saving increases by ₹ 40 more than the saving of immediately previous month. His total saving from the start of service will be ₹ 11040 after
A 19 months
B 20 months
C 21 months
D 18 months
Correct Answer
Option C
Solution

Let required number of months

=n=n

∴\therefore

200×3+(240+280+320+...200 \times 3 + \left( {240 + 280 + 320 + ...} \right.
            +(n−3)th term)=11040\left. {\,\,\,\,\,\,\,\,\,\,\,\, + {{\left( {n - 3} \right)}^{th}}\,term} \right) = 11040
⇒n−32[2×240+(n−4)×40]\Rightarrow {{n - 3} \over 2}\left[ {2 \times 240 + \left( {n - 4} \right) \times 40} \right]
            =11040−600\,\,\,\,\,\,\,\,\,\,\,\, = 11040 - 600
⇒(n−3)[240+20n−80]=10440\Rightarrow \left( {n - 3} \right)\left[ {240 + 20n - 80} \right] = 10440
⇒(n−3)(20n+160)=10440\Rightarrow \left( {n - 3} \right)\left( {20n + 160} \right) = 10440
⇒(n−3)(n+8)=522\Rightarrow \left( {n - 3} \right)\left( {n + 8} \right) = 522
⇒n2+5n−546=0\Rightarrow {n^2} + 5n - 546 = 0
⇒(n+26)(n−21)=0\Rightarrow \left( {n + 26} \right)\left( {n - 21} \right) = 0

∴\therefore

n=21n = 21
Q106
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 :
A 12110{{121} \over {10}}
B -12110{{121} \over {10}}
C 725{{72} \over 5}
D -725{{72} \over 5}
Correct Answer
Option D
Solution

Let common difference be d. ∵\because a1 + a2 + a3 + ... + a11 = 0 ∴\therefore

112[2a1+10d]{{11} \over 2}\left[ {2{a_1} + 10d} \right]

= 0 ⇒\Rightarrow a1 + 5d = 0 ⇒\Rightarrow d =

−a15{ - {{{a_1}} \over 5}}

.....(1) Now a1 + a3 + a5 + ... + a23 = (a1 + a23) ×\times

122{{12} \over 2}

= (a1 + a1 + 22d) × 6 =

[2a1+22(−a15)]\left[ {2{a_1} + 22\left( { - {{{a_1}} \over 5}} \right)} \right]

×\times 6 =

−722a1- {{72} \over 2}{a_1}

∴\therefore k =

−722- {{72} \over 2}
Q107
Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d. If the fourth term of GP is 3 r2, then r2 −- d is equal to :
A 7 −- 73\sqrt 3
B 7 + 3\sqrt 3
C 7 −- 3\sqrt 3
D 7 + 33\sqrt 3
Correct Answer
Option B
Solution

Let numbers be

ar{a \over r}

, a, ar →\to G.P.

ar{a \over r}

, 2a, ar →\to A.P. ⇒\Rightarrow 4a =

ar{a \over r}

+ ar ⇒\Rightarrow r +

1r{1 \over r}

= 4 r = 2 ±\pm

3\sqrt 3

4th form of G.P. = 3r2 ⇒\Rightarrow ar2 = 3r2 ⇒\Rightarrow a = 3 r = 2 +

3\sqrt 3

, a = 3, d = 2a −-

ar{a \over r}

= 3

3\sqrt 3

r2 −- d = (2 +

3\sqrt 3

)2 −- 3

3\sqrt 3

= 7 + 4

3\sqrt 3

−- 3

3\sqrt 3

= 7 +

3\sqrt 3
Q108
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) + ....
A x+y−xy(1+x)(1+y){{x + y - xy} \over {\left( {1 + x} \right)\left( {1 + y} \right)}}
B x+y−xy(1−x)(1−y){{x + y - xy} \over {\left( {1 - x} \right)\left( {1 - y} \right)}}
C x+y+xy(1+x)(1+y){{x + y + xy} \over {\left( {1 + x} \right)\left( {1 + y} \right)}}
D x+y+xy(1−x)(1−y){{x + y + xy} \over {\left( {1 - x} \right)\left( {1 - y} \right)}}
Correct Answer
Option B
Solution

(x + y) + (x2+xy+y2) + (x3+x2y + xy2+y3) + .... By multiplying and dividing x – y :

(x2−y2)+(x3−y3)+(x4−y4)+...x−y{{\left( {{x^2} - {y^2}} \right) + \left( {{x^3} - {y^3}} \right) + \left( {{x^4} - {y^4}} \right) + ...} \over {x - y}}

=

(x2+x3+x4+....)−(y2+y3+y4+...)x−y{{\left( {{x^2} + {x^3} + {x^4} + ....} \right) - \left( {{y^2} + {y^3} + {y^4} + ...} \right)} \over {x - y}}

=

x21−x−y21−yx−y{{{{{x^2}} \over {1 - x}} - {{{y^2}} \over {1 - y}}} \over {x - y}}

=

(x2−y2)−xy(x−y)(1−x)(1−y)(x−y){{\left( {{x^2} - {y^2}} \right) - xy\left( {x - y} \right)} \over {\left( {1 - x} \right)\left( {1 - y} \right)\left( {x - y} \right)}}

=

x+y−xy(1−x)(1−y){{x + y - xy} \over {\left( {1 - x} \right)\left( {1 - y} \right)}}
Q109
The sum of the series ∑n=1∞n2+6n+10(2n+1)!\sum\limits_{n = 1}^\infty {{{{n^2} + 6n + 10} \over {(2n + 1)!}}} is equal to :
A 418e+198e−1−10{{41} \over 8}e + {{19} \over 8}{e^{ - 1}} - 10
B 418e−198e−1−10{{41} \over 8}e - {{19} \over 8}{e^{ - 1}} - 10
C 418e+198e−1+10{{41} \over 8}e + {{19} \over 8}{e^{ - 1}} + 10
D −418e+198e−1−10 - {{41} \over 8}e + {{19} \over 8}{e^{ - 1}} - 10
Correct Answer
Option B
Solution
∑n=1∞n2+6n+10(2n+1)!\sum\limits_{n = 1}^\infty {{{{n^2} + 6n + 10} \over {(2n + 1)!}}}

Put 2n + 1 = r, where r = 3, 5, 7, .......

⇒n=r−12\Rightarrow n = {{r - 1} \over 2}
n2+6n+10(2n+1)!=(r−12)2+3r−3+10r!{{{n^2} + 6n + 10} \over {(2n + 1)!}} = {{{{\left( {{{r - 1} \over 2}} \right)}^2} + 3r - 3 + 10} \over {r!}}
=r2+10r+294r!= {{{r^2} + 10r + 29} \over {4r!}}

=

r(r−1)+11r+294r!{{{r(r - 1) + 11r + 29} \over {4r!}}}

Now,

∑r=3,5,7r(r−1)+11r+294r!\sum\limits_{r = 3,5,7} {{{r(r - 1) + 11r + 29} \over {4r!}}}

=

14∑r=3,5,7,......(1(r−2)!+11(r−1)!+29r!){1 \over 4}\sum\limits_{r = 3,5,7,......} {\left( {{1 \over {(r - 2)!}} + {{11} \over {(r - 1)!}} + {{29} \over {r!}}} \right)}
=14{(11!+13!+15!+......)+11(12!+14!+16!+......)+29(13!+15!+17!+......)}= {1 \over 4}\left\{ {\left( {{1 \over {1!}} + {1 \over {3!}} + {1 \over {5!}} + ......} \right) + 11\left( {{1 \over {2!}} + {1 \over {4!}} + {1 \over {6!}} + ......} \right) + 29\left( {{1 \over {3!}} + {1 \over {5!}} + {1 \over {7!}} + ......} \right)} \right\}
=14{e−1e2+11(e+1e−22)+29(e−1e−22)}= {1 \over 4}\left\{ {{{e - {1 \over e}} \over 2} + 11\left( {{{e + {1 \over e} - 2} \over 2}} \right) + 29\left( {{{e - {1 \over e} - 2} \over 2}} \right)} \right\}
=18{e−1e+11e+11e−22+29e−29e−58}= {1 \over 8}\left\{ {e - {1 \over e} + 11e + {{11} \over e} - 22 + 29e - {{29} \over e} - 58} \right\}
=18{41e−19e−80}= {1 \over 8}\left\{ {41e - {{19} \over e} - 80} \right\}
Q110
The common difference of the A.P. b1, b2, … , bm is 2 more than the common difference of A.P. a1, a2, …, an. If a40 = –159, a100 = –399 and b100 = a70, then b1 is equal to :
A 127
B 81
C –127
D -81
Correct Answer
Option D
Solution

Let common difference of series a1 , a2 , a3 ,..., an be d. ∵\because a40 = a1 + 39d == –159 ...(i) and a100 = a1 + 99d = –399 ...(ii) From eqn. (ii) and (i) d = –4 and a1 = –3.

The common difference of the A.P. b1, b2, … , bm is 2 more than the common difference of A.P. a1, a2, …, an.

∴\therefore Common difference of b1 , b2 , b3 , ..., be (–2). ∵\because b100 = a70 ∴\therefore b1 + 99(–2) = (–3) + 69(–4) ∴\therefore b1 = 198 – 279 ∴\therefore b1 = – 81

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