Sequences and Series — Page 17 | JEE Mathematics

Q161

The sum of all natural numbers 'n' such that 100 < n < 200 and H.C.F. (91, n) > 1 is :

A 3221

B 3121

C 3203

D 3303

Correct Answer

Option B

Solution

$\because$ 91 = 13 $\times$ 7 So the required numbers are either divisible by 7 or 13.

SA = sum of numbers between 100 and 200 which are divisible by 7.

$\Rightarrow$ SA = 105 + 112 + .....

+ 196 SA =

{{14} \over 2}\left[ {105 + 196} \right]

= 2107 SB = Sum of numbers between 100 and 200 which are divisible by 13. SB = 104 + 117 + ....... + 195 SB =

{8 \over 2}\left[ {104 + 195} \right]

= 1196 SC = Sum of numbers between 100 and 200 which are divisible by 7 and 13.

SC = 182 Sum of numbers divisible by 7 or 13 = Sum of no. divisible by 7 + sum of the no. divisible by 13 – Sum of the numbers divisible by 7 and 13 = 2107 + 1196 - 182 = 3121

Q162

For x

\varepsilon

R, let [x] denote the greatest integer

\le

x, then the sum of the series

\left[ { - {1 \over 3}} \right] + \left[ { - {1 \over 3} - {1 \over {100}}} \right] + \left[ { - {1 \over 3} - {2 \over {100}}} \right] + .... + \left[ { - {1 \over 3} - {{99} \over {100}}} \right]

is :

A - 153

B - 135

C - 133

D - 131

Correct Answer

Option C

Solution

\left[ {{{ - 1} \over 3}} \right] + \left[ {{{ - 1} \over 3} - {1 \over {100}}} \right] + \left[ {{{ - 1} \over 3} - {2 \over {100}}} \right] +

....... + \left[ {{{ - 1} \over 3} - {{99} \over {100}}} \right]

\Rightarrow ( - 1 - 1 - 1 - .....67\,\times) + ( - 2 - 2 - 2 - .....33\,\times)

$\Rightarrow$ -133

Q163

Let a1, a2, a3,......be an A.P. with a6 = 2. Then the common difference of this A.P., which maximises the product a1a4a5, is :

A

{3 \over 2}

B

{6 \over 5}

C

{8 \over 5}

D

{2 \over 3}

Correct Answer

Option C

Solution

first term = a, Common difference = d $\therefore$ a + 5d = 2 a1. a4. a5 = a(a + 3d) (a + 4d) f(d) = (2 – 5d) (2 – 2d) (2 – d) $\Rightarrow$

f'(d) = 0 \Rightarrow d = {2 \over 3},{8 \over 5}

$\Rightarrow$

f''(d) < 0\,at\,d = {8 \over 5}

\, \Rightarrow d = {8 \over 5}

Q164

The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is -

A 1356

B 1256

C 1365

D 1465

Correct Answer

Option A

Solution

\sum\limits_{r = 2}^{13} {(7r + 2) = 7.{{2 + 13} \over 2}} \times 6 + 2 \times 12

= 7 $\times$ 90 + 24 = 654

\sum\limits_{r = 1}^{13} {(7r + 5) = 7\left( {{{1 + 13} \over 2}} \right)} \times 13 + 5 \times 13 = 702

Total = 654 + 702 = 1356

Q165

If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P. is :

A

{1 \over 4}

B

{1 \over 5}

C

{1 \over 7}

D

{1 \over 6}

Correct Answer

Option D

Solution

First 25 terms = a, a + d, .......,a + 24d Next 15 terms = a + 25d, a + 26d, ......, a + 39d $\therefore$

{{25} \over 2}\left[ {2a + 24d} \right] = {{15} \over 2}\left[ {2\left( {a + 25d} \right) + 14d} \right]

$\Rightarrow$ 50a + 600d = 15 [2a + 50d + 14d] $\Rightarrow$ 20a + 600d = 960d $\Rightarrow$ 60 = 360d $\Rightarrow$ d =

{1 \over 6}

Q166

If

{3^{2\sin 2\alpha - 1}}

, 14 and

{3^{4 - 2\sin 2\alpha }}

are the first three terms of an A.P. for some

\alpha

, then the sixth terms of this A.P. is:

A 66

B 81

C 65

D 78

Correct Answer

Option A

Solution

Given that

{3^{4 - \sin 2\alpha }} + {3^{2\sin 2\alpha - 1}} = 28

Let

{3^{2\sin 2\alpha }}

= t $\Rightarrow$

{{81} \over t} + {t \over 3} = 28

$\Rightarrow$ t = 81, 3 $\therefore$

{3^{2\sin 2\alpha }}

= 31, 34

\sin 2\alpha = {1 \over 2}

, 2 (rejected) First term a =

{3^{2\sin 2\alpha -1}}

= 30 $\Rightarrow$ a = 1 Given Second term = 14 $\therefore$ Common difference d = 13

{T_6} = a + 5d

{T_6} = 1 + 5 \times 13

{T_6} = 66

Q167

If the sum of the series 20 + 19

{3 \over 5}

+ 19

{1 \over 5}

+ 18

{4 \over 5}

+ ... upto nth term is 488 and the nth term is negative, then :

A n = 41

B n = 60

C nth term is –4

D nth term is -4

{2 \over 5}

Correct Answer

Option C

Solution

S = {{100} \over 5} + {{98} \over 5} + {{96} \over 5} + {{94} \over 5} + ...\,n

\,{S_n} = {n \over 2}\left( {2 \times {{100} \over 5} + (n - 1)\left( {{{ - 2} \over 5}} \right)} \right) = 488

$\Rightarrow$

n(100 - n + 1) = 488 \times 5

$\Rightarrow$

{n^2} - 101n + 488 \times 5 = 0

$\Rightarrow$

n = 61,\,40

For negative term n = 61

\,{T_n} = a + (n - 1)d = {{100} \over 5} - {2 \over 5} \times 60

= 20 - 24 = - 4

Q168

Let

{a_1}

,

{a_2}

,

{a_3}

,....... be a G.P. such that

{a_1}

< 0,

{a_1}

+

{a_2}

= 4 and

{a_3}

+

{a_4}

= 16. If

\sum\limits_{i = 1}^9 {{a_i}} = 4\lambda

, then

\lambda

is equal to:

A 171

B -171

C -513

D

{{511} \over 3}

Correct Answer

Option B

Solution

{a_1}

+

{a_2}

= 4 $\Rightarrow$

{a_1}

+

{a_1}

r = 4 ...(1)

{a_3}

+

{a_4}

= 16 $\Rightarrow$

{a_1}

r2 +

{a_1}

r3 = 16 ...(2) Doing (1)

\div

(2), we get r = $\pm$ 2 If r = 2, then a1 =

{4 \over 3}

If r = -2, then a1 = -4 Given

{a_1}

< 0 $\therefore$ a1 = -4 $\therefore$

\sum\limits_{i = 1}^9 {{a_i}}

=

{{a\left( {{r^9} - 1} \right)} \over {r - 1}}

= 4 $\lambda$ $\Rightarrow$

{{ - 4\left( {{{\left( { - 2} \right)}^9} - 1} \right)} \over { - 2 - 1}}

= 4 $\lambda$ $\Rightarrow$ $\lambda$ = -171

Q169

If the sum of the first 40 terms of the series, 3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + ..... is (102)m, then m is equal to :

A 20

B 5

C 10

D 25

Correct Answer

Option A

Solution

3 + 4 + 8 + 9 + 13 + 14 +…….upto 40 terms $\Rightarrow$ 7 + 17 + 27 +…….20 terms S =

{{20} \over 2}

[2 × 7 + 19 × 10] = 102 × 20 = 102 m $\therefore$ m = 20

Q170

Let ƒ : R

\to

R be such that for all x

\in

R (21+x + 21–x), ƒ(x) and (3x + 3–x) are in A.P., then the minimum value of ƒ(x) is

A 2

B 0

C 3

D 4

Correct Answer

Option C

Solution

f(x) =

{{2\left( {{2^x} + {2^{ - x}}} \right) + \left( {{3^x} + {3^{ - x}}} \right)} \over 2} \ge 3

As we know, A.M > G.M