Sequences and Series — Page 13 | JEE Mathematics

Q121

For

0 (I) If

\alpha \in(-1,0)

, then

b

cannot be the geometric mean of a and

c

(II) If

\alpha \in(0,1)

, then

b

may be the geometric mean of

a

and

c$$

A only (II) is true

B Both (I) and (II) are true

C only (I) is true

D Neither (I) nor (II) is true

Correct Answer

Option B

Solution

\begin{aligned} & f(x)=(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b) \\ & f(x)=a+b-2 c+b+c-2 a+c+a-2 b=0 \\ & f(1)=0 \\ & \therefore \alpha \cdot 1=\frac{c+a-2 b}{a+b-2 c} \\ & \alpha=\frac{c+a-2 b}{a+b-2 c} \\ & \text{ If, }-1\frac{a+c}{2} \end{aligned}

therefore, b cannot be G.M. between a and c.

\begin{aligned} & \text{ If, } 0c \text{ and } b Therefore,

\mathrm{b}

may be the G.M. between

\mathrm{a}

and

\mathrm{c}$$.

Q122

For

x \geqslant 0

, the least value of

\mathrm{K}

, for which

4^{1+x}+4^{1-x}, \dfrac{\mathrm{K}}{2}, 16^x+16^{-x}

are three consecutive terms of an A.P., is equal to :

A 10

B 4

C 8

D 16

Correct Answer

Option A

Solution

To determine the least value of

\mathrm{K}

for which the terms

4^{1+x} + 4^{1-x}, \frac{\mathrm{K}}{2}, 16^x + 16^{-x}

form an arithmetic progression (A.P.), we need to establish the relationship among these terms in an A.P.

For three numbers to be in an arithmetic progression, the middle term must be the average of the other two terms.

Therefore, we can write:

\frac{4^{1+x} + 4^{1-x} + 16^x + 16^{-x}}{2} = \frac{K}{2}

First, simplify each term individually: 1. Consider

4^{1+x} + 4^{1-x}

:

4^{1+x} = 4 \cdot 4^x = 4 \cdot (2^2)^x = 4 \cdot 2^{2x} = 4 \cdot 2^{2x}

and

4^{1-x} = 4 \cdot 4^{-x} = 4 \cdot (2^2)^{-x} = 4 \cdot 2^{-2x} = 4 \cdot 2^{-2x}

Thus,

4^{1+x} + 4^{1-x} = 4 \cdot 2^{2x} + 4 \cdot 2^{-2x} = 4(2^{2x} + 2^{-2x})

2. Consider

16^x + 16^{-x}

:

16^x = (2^4)^x = 2^{4x}

and

16^{-x} = (2^4)^{-x} = 2^{-4x}

Thus,

16^x + 16^{-x} = 2^{4x} + 2^{-4x}

3. Combine the terms and set up the equation:

\frac{4(2^{2x} + 2^{-2x}) + 2^{4x} + 2^{-4x}}{2} = \frac{K}{2}

Multiply both sides by 2:

4(2^{2x} + 2^{-2x}) + 2^{4x} + 2^{-4x} = K

To find the least value of

\mathrm{K}

, let's assume

x = 0

(since

x

can range over non-negative values): For

x = 0

:

4(2^{2 \cdot 0} + 2^{-2 \cdot 0}) + 2^{4 \cdot 0} + 2^{-4 \cdot 0}

This simplifies to:

4(2^0 + 2^0) + 2^0 + 2^0

= 4(1 + 1) + 1 + 1

= 4 \cdot 2 + 1 + 1

= 8 + 1 + 1

= 10

Therefore, the least value of

\mathrm{K}

that ensures the values form an arithmetic progression is $10$ . Hence, the correct option is: Option A: 10

Q123

A software company sets up m number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of

\mathrm{m}

is equal to:

A 125

B 160

C 150

D 180

Correct Answer

Option C

Solution

To determine the value of

\mathrm{m}

, we need to formulate the problem using some basic concepts of arithmetic progression and work.

Let's first understand the nature of the problem: Initially, there are

\mathrm{m}

computers, and it is estimated that with these

\mathrm{m}

computers, the assignment can be completed in 17 days.

However, due to the crash of 4 computers every day starting from the second day onward, the total time taken extends by 8 days, making it 25 days in total.

To begin with, let's define the total work (W) in terms of the number of computers and days: The total work (W) is given by: The amount of work completed each day with

\mathrm{m}

computers for 17 days:

W = 17m

When computers crash, the number of working computers each day forms an arithmetic sequence. On the first day, there are

\mathrm{m}

computers. On the second day, there are

\mathrm{m} - 4

computers, on the third day, there are

\mathrm{m} - 8

computers, and so on.

We need to sum this series until 25 days are completed.

This can be formulated as: Total work done over 25 days with decrement in the number of computers:

W = m + (m - 4) + (m - 8) + \ldots + \left[m - 4 \times (n - 1)\right]

where

n

is the number of days. Here,

n = 25

. Notice that we form an arithmetic series where the first term (a) is

\mathrm{m}

and the common difference (d) is -4. The sum of the first n terms of an arithmetic series is:

S_n = \frac{n}{2} \left[ 2a + (n - 1)d \right]

Plugging in the values:

S_{25} = \frac{25}{2} \left[ 2m + (25 - 1)(-4) \right]

S_{25} = \frac{25}{2} \left[ 2m - 96 \right]

S_{25} = \frac{25}{2} \left[ 2m - 96 \right] = 25(m - 48)

This work should be equivalent to the work calculated earlier, so:

17m = 25(m - 48)

Solving for

\mathrm{m}

:

17m = 25m - 1200

8m = 1200

m = 150

Thus, the value of

\mathrm{m}

is equal to: Option C: 150

Q124

The sum of the first 20 terms of the series

1 + {3 \over 2} + {7 \over 4} + {{15} \over 8} + {{31} \over {16}} + ...,

is :

A

38 + {1 \over {{2^{19}}}}

B

38 + {1 \over {{2^{20}}}}

C

39 + {1 \over {{2^{20}}}}

D

39 + {1 \over {{2^{19}}}}

Correct Answer

Option A

Solution

1 +

{3 \over 2}

+

{7 \over 4}

+

{15 \over 8}

+

{31 \over 16}

+ . . . . = (2 $-$ 1) + (2 $-$

{1 \over 2}

) + (2 $-$

{1 \over 4}

) + (2 $-$

{1 \over 8}

) + . . . . .+ 20 terms = (2 + 2 + . . . . . 20 terms) $-$ (1 +

{1 \over 2}

+

{1 \over 4}

+ . . . . . 20 terms) = 2 $\times$ 20 $-$

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

= 40 $-$ 2 + 2

{\left( {{1 \over 2}} \right)^{20}}

= 38 +

{1 \over {{2^{19}}}}

Q125

If a, b, c be three distinct real numbers in G.P. and a + b + c = xb , then x cannot be

A 2

B -3

C 4

D -2

Correct Answer

Option A

Solution

a, b, c are in G.P.

So, b = ar and c = ar2 given a + b + c = xb $\Rightarrow$ a + br + ar2 = x(ar) $\Rightarrow$ 1 + r + r2 = xr $\Rightarrow$ x = 1 + r +

{1 \over r}

let sum of r +

{1 \over r}

= M $\therefore$ r2 + 1 = Mr $\Rightarrow$ r2 $-$ Mr + 1 = 0 this quadratic equation will have real solution when discriminant is $\ge$ 0 $\therefore$ b2 $-$ 4ac $\ge$ 0 M2 $-$ 4.1.1 $\ge$ 0 $\Rightarrow$ M2 $\ge$ 4 M $\ge$ 2 or M $\le$ $-$ 2 $\therefore$ M

\in

( $-$ $\propto$ , $-$ 2] $\cup$ [2, $\propto$ ) As x = 1 + r +

{1 \over r}

= 1 + M $\therefore$ x

\in

( $-$ $\propto$ , $-$ 1] $\cup$ [3, $\propto$ ) $\therefore$ x can't be 0, 1, 2.

Q126

Let

{a_1},{a_2},.......,{a_{30}}

be an A.P.,

S = \sum\limits_{i = 1}^{30} {{a_i}}

and

T = \sum\limits_{i = 1}^{15} {{a_{\left( {2i - 1} \right)}}}

. If

a_5

= 27 and S - 2T = 75, then

a_{10}

is equal to :

A 47

B 42

C 52

D 57

Correct Answer

Option C

Solution

Let the common difference = d S =

\sum\limits_{i = 1}^{30} {{a_i}}

=

a

1 +

a

2 + . . . . . +

a

30 $\therefore$ S =

{{30} \over 2}\left[ {{a_1} + {a_{30}}} \right]

= 15 [

a

1 +

a

1 + 29d] = 15 (2

a

1 + 29d) T =

\sum\limits_{i = 1}^{15} {{a_{\left( {2i - 1} \right)}}}

=

a

1 +

a

3 + . . . . . . +

a

29 =

{{15} \over 2}\left[ {a{}_1 + {a_{29}}} \right]

=

{{15} \over 2}\left[ {a{}_1 + {a_1} + 28d} \right]

=

{{15} \over 2}\left[ {2a{}_1 + 28d} \right]

= 15 (

a

1 + 14d) Given, S $-$ 2T = 75 $\Rightarrow$ 15(2

a

1 + 29d) $-$ 2 $\times$ 15 (

a

1 + 14d) = 75 $\Rightarrow$ 30

a

1 + 15 $\times$ 29d $-$ 30

a

1 $-$ 420d = 75 $\Rightarrow$ 435d $-$ 420d = 75 $\Rightarrow$ 15d = 75 $\Rightarrow$ d = 5 Given that,

a

5 = 27 $\Rightarrow$

a

1 + 4d = 27 $\Rightarrow$

a

1 + 20 = 27 $\Rightarrow$

a

1 = 7 $\therefore$

a

10 =

a

1 + 9d = 7 + 45 = 52

Q127

The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an A.P. Then the sum of the original three terms of the given G.P. is :

A 36

B 28

C 32

D 24

Correct Answer

Option B

Solution

Let terms are

{a \over r},a,ar \to G.P

$\therefore$

{a^3}

= 512 $\Rightarrow$ a = 8

{8 \over r} + 4,12,8r \to A.P.

24 =

{8 \over r} + 4 + 8r

r = 2, r =

{1 \over 2}

r = 2(4, 8, 16) r =

{1 \over 2}

(16, 8, 4) Sum = 28

Q128

Let S be the sum of the first 9 terms of the series : {x + k

a

} + {x2 + (k + 2)

a

} + {x3 + (k + 4)

a

} + {x4 + (k + 6)

a

} + .... where a

\ne

0 and x

\ne

1. If S =

{{{x^{10}} - x + 45a\left( {x - 1} \right)} \over {x - 1}}

, then k is equal to :

A -3

B 1

C -5

D 3

Correct Answer

Option A

Solution

S = {x + k

a

} + {x2 + (k + 2)

a

} + {x3 + (k + 4)

a

} + {x4 + (k + 6)

a

} + ....

S = \left( {x + {x^2} + {x^3} + ....\,9terms} \right) +

a\left( {k + \left( {k + 2} \right) + (k + 4) + (k + 6) + ....9terms} \right)

S = {{x\left( {{x^9} - 1} \right)} \over {\left( {x - 1} \right)}} + a\left[ {{9 \over 2}\left[ {2k + \left( {9 - 1} \right)2} \right]} \right]

S = {{{x^{10}} - x} \over {x - 1}} + 9a\left( {k + 8} \right)

S = {{{x^{10}} - x + 9a\left( {k + 8} \right)\left( {x - 1} \right)} \over {x - 1}}

S = {{{x^{10}} - x + 9\left( {k + 8} \right)a\left( {x - 1} \right)} \over {x - 1}}

Compare with given sum, then we get

{9\left( {k + 8} \right) = 45}

\Rightarrow \left( {k + 8} \right) = 5

\Rightarrow k = - 3

Q129

\begin{aligned} &\text{ Let }\left\{a_{n}\right\}_{n=0}^{\infty} \text{ be a sequence such that } a_{0}=a_{1}=0 \text{ and } \\\\ &a_{n+2}=3 a_{n+1}-2 a_{n}+1, \forall n \geq 0 . \end{aligned}

Then

a_{25} a_{23}-2 a_{25} a_{22}-2 a_{23} a_{24}+4 a_{22} a_{24}

is equal to

A 483

B 528

C 575

D 624

Correct Answer

Option B

Solution

Given,

{a_0} = {a_1} = 0

and

{a_{n + 2}} = 3{a_{n + 1}} - 2{a_n} + 1

For

n = 0,\,{a_2} = 3{a_1} - 2{a_0} + 1

= 3\,.\,0 - 2\,.\,0 + 1

= 1

For

n = 1,\,{a_3} = 3{a_2} - 2{a_1} + 1

= 3\,.\,1 - 2\,.\,0 + 1

= 4

For

n = 2,\,{a_4} = 3{a_3} - 2{a_2} + 1

= 3\,.\,4 - 2\,.\,1 + 1

= 11

For

n = 3,\,{a_5} = 3{a_4} - 2{a_3} + 1

= 3\,.\,11 - 2\,.\,4 + 1

= 26

For

n = 4,\,{a_6} = 3{a_5} - 2{a_4} + 1

= 3\,.\,26 - 2\,.\,11 + 1

= 57

$\therefore$

{S_n} = 1 + 4 + 11 + 26 + 57\, + \,....\, + \,{t_n}

{S_n} = 1 + 4 + 11 + 26\, + \,....\, + \,{t_{n - 1}} + {t_n}

0 = 1 + 3 + 7 + 15 + 31\, + \,.....\, - {t_n}

\Rightarrow {t_n} = 1 + 3 + 7 + 15 + 31\, + \,....

Now, find the sum of the series,

{t_n} = 1 + 3 + 7 + 15 + 31\, + \,.....\, + \,{x_{n - 1}} + {x_n}

.....(1)

{t_n} =

1 + 3 + 7 + 15\, + \,.....\, + \,{x_{n - 1}} + {x_n}

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

0 = 1 + 2 + 4 + 8 + 16\, + \,....\, + \,{x_n}

\Rightarrow {x_n} = 1 + 2 + 4 + 8 + 16\, + \,.....\, + \,

n terms

= {{1({2^n} - 1)} \over {2 - 1}}

= {2^n} - 1

$\therefore$

{t_n} = \sum\limits_{n = 1}^n {{x_n}}

= \sum\limits_{n = 1}^n {({2^n} - 1)}

= \sum\limits_{n = 1}^n {{2^n} - \sum\limits_{n = 1}^n 1 }

= {{2({2^n} - 1)} \over {2 - 1}} - n

= {2^{n + 1}} - 2 - n

{t_1} = {2^2} - 2 - 1 = 1 = {a_2}

{t_2} = {2^3} - 2 - 2 = 4 = {a_3}

{t_3} = {2^4} - 2 - 3 = 11 = {a_4}

$\therefore$

{a_{22}} = {t_{21}} = {2^{22}} - 2 - 21 = {2^{22}} - 23

{a_{23}} = {t_{22}} = {2^{23}} - 2 - 22 = {2^{23}} - 24

{a_{24}} = {t_{23}} = {2^{24}} - 2 - 23 = {2^{24}} - 25

{a_{25}} = {t_{24}} = {2^{25}} - 2 - 24 = {2^{25}} - 26

Now,

{a_{25}}{a_{23}} - 2{a_{25}}{a_{22}} - 2{a_{23}}{a_{24}} + 4{a_{22}} {a_{24}}

= {a_{25}}({a_{23}} - 2{a_{22}}) - 2{a_{24}}({a_{23}} - 2{a_{22}})

= ({a_{23}} - 2{a_{22}})({a_{25}} - 2{a_{24}})

= [({2^{23}} - 24) - 2({2^{22}} - 23)][({2^{25}} - 26) - 2({2^{24}} - 25)]

= [({2^{23}} - 24 - {2^{23}} + 46)][({2^{25}} - 26 - {2^{25}} + 50)]

= (22)(24)

= 528

Q130

If the sum of the first n terms of the series

\,\sqrt 3 + \sqrt {75} + \sqrt {243} + \sqrt {507} + ......

is

435\sqrt 3 ,

then n equals :

A 18

B 15

C 13

D 29

Correct Answer

Option B

Solution

Given,

\sqrt 3

+

\sqrt {75}

+

\sqrt {243}

+

\sqrt {507}

+ . . . . . .+ n terms =

\sqrt 3

+

\sqrt {25 \times 3}

+

\sqrt {81 \times 3}

+

\sqrt {169 \times 3}

+ . . . . . .+ n terms =

\sqrt 3

+ 5

\sqrt 3

+ 9

\sqrt 3

+ 13

\sqrt 3

+ . . . . . .+ n terms =

\sqrt 3

[ 1 + 5 + 9 + 13 + . . . . .+ n terms] =

\sqrt 3

\left[ {{n \over 2}\left( {2.1 + \left( {n - 1} \right)4} \right)} \right]

=

\sqrt 3

\left[ {{n \over 2}\left( {2 + 4n - 4} \right)} \right]

=

\sqrt 3

\left[ {{n \over 2}\left( {4n - 2} \right)} \right]

=

\sqrt 3

[n (2n $-$ 1)] According to question,

\sqrt 3

[n (2n $-$ 1)] = 435

\sqrt 3

$\Rightarrow$

\,\,\,

2n2 $-$ n = 435

\therefore\,\,\,

n =

{{1 \pm \sqrt {1 + 4 \times 2 \times 435} } \over 4}

=

{{1 \pm 59} \over 4}

\therefore\,\,\,

n =

{{1 + 59} \over 4}

= 15 or

{{1 - 59} \over 4}

= $-$ 14.5

\therefore\,\,\,

n = 15 (as n can't be $-$ ve)