Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

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General Aptitude

1

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

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$$\sqrt 3 $$

$$ \Rightarrow $$$$\,\,\,$$ 2n^{2} $$-$$ 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)

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

$$ \Rightarrow $$$$\,\,\,$$ 2n

$$\therefore\,\,\,$$ 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)

2

If three positive numbers a, b and c are in A.P. such that abc = 8, then the minimum possible value of b is :

A

2

B

4$${^{{1 \over 3}}}$$

C

4$${^{{2 \over 3}}}$$

D

4

a, b and c are in AP.

$$ \therefore $$ a + c = 2b

As, abc = 8

$$ \Rightarrow $$ac$$\left( {{{a + c} \over 2}} \right)$$= 8

$$ \Rightarrow $$ ac(a + c) = 16 = 4 $$ \times $$ 4

$$ \therefore $$ ac = 4 and a + c = 4

Then,

b = $$\left( {{{a + c} \over 2}} \right)$$ = $${4 \over 2}$$ = 2

$$ \therefore $$ a + c = 2b

As, abc = 8

$$ \Rightarrow $$ac$$\left( {{{a + c} \over 2}} \right)$$= 8

$$ \Rightarrow $$ ac(a + c) = 16 = 4 $$ \times $$ 4

$$ \therefore $$ ac = 4 and a + c = 4

Then,

b = $$\left( {{{a + c} \over 2}} \right)$$ = $${4 \over 2}$$ = 2

3

Let

S_{n} = $${1 \over {{1^3}}}$$$$ + {{1 + 2} \over {{1^3} + {2^3}}} + {{1 + 2 + 3} \over {{1^3} + {2^3} + {3^3}}} + ......... + {{1 + 2 + ....... + n} \over {{1^3} + {2^3} + ...... + {n^3}}}.$$

If 100 S_{n} = n, then n is equal to :

S

If 100 S

A

199

B

99

C

200

D

19

n^{th} term, T_{n} = $${{1 + 2 + .... + n} \over {{1^2} + {2^2} + .... + {n^2}}}$$

T_{n} = $${{{{n\left( {n + 1} \right)} \over 2}} \over {{{\left( {{{n\left( {n + 1} \right)} \over 2}} \right)}^2}}}$$

$$ \Rightarrow $$ T_{n} = $${2 \over {n\left( {n + 1} \right)}}$$ = $$2\left[ {{1 \over n} - {1 \over {n + 1}}} \right]$$

$$ \therefore $$ S_{n} = $$\sum {{T_n}} $$

= $$2\sum\limits_{n = 1}^n {\left[ {{1 \over n} - {1 \over {n + 1}}} \right]} $$

= $$2\left( {1 - {1 \over n}} \right)$$

= $${{{2n} \over {n + 1}}}$$

Given that,

100 S_{n} = n

$$ \Rightarrow $$ 100 $$ \times $$ $${{{2n} \over {n + 1}}}$$ = n

$$ \Rightarrow $$ n + 1 = 200

$$ \Rightarrow $$ n = 199

T

$$ \Rightarrow $$ T

$$ \therefore $$ S

= $$2\sum\limits_{n = 1}^n {\left[ {{1 \over n} - {1 \over {n + 1}}} \right]} $$

= $$2\left( {1 - {1 \over n}} \right)$$

= $${{{2n} \over {n + 1}}}$$

Given that,

100 S

$$ \Rightarrow $$ 100 $$ \times $$ $${{{2n} \over {n + 1}}}$$ = n

$$ \Rightarrow $$ n + 1 = 200

$$ \Rightarrow $$ n = 199

4

Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series

1^{2} + 2.2^{2} + 3^{2} + 2.4^{2} + 5^{2} + 2.6^{2} ...........

If B - 2A = 100$$\lambda $$, then $$\lambda $$ is equal to

1

If B - 2A = 100$$\lambda $$, then $$\lambda $$ is equal to

A

496

B

232

C

248

D

464

Sum of square of first n odd terms

1

Given,

1

A = Sum of first 20 terms

$$\therefore\,\,\,$$A = 1

Arrange those terms this way,

A = [1

A = [ 1

A = $$ {{10 \times \left( {2.10 - 1} \right)\left( {2.10 + 1} \right)} \over 3} + {2.2^2}\left[ {{{10 \times 11 \times 21} \over 6}} \right]$$

A = $$ {{10 \times 19 \times 21} \over 3} + 8 \times {{10 \times 11 \times 21} \over 6}$$

A =70 $$ \times $$ 19 + 70 $$ \times $$ 44

A = 70 $$ \times $$ 63

B = Sum of first 40 terms

Arrange those terms this way.

B = [1

B = [1

B = $${{20 \times 39 \times 41} \over 3} + \,\,8\,\, \times {{20 \times 21 \times 41} \over 6}$$

B = 260 $$ \times $$ 41 + 560 $$ \times $$ 41

B = 41 $$ \times \,\,\,820$$

$$\therefore\,\,\,$$ B $$-$$ 2A = 41 $$ \times \,$$ 820 $$-$$ 2 $$ \times \,$$ 70 $$ \times \,$$ 63 = 24800

Given that B $$-$$ 2A = 100 $$\lambda $$

$$\therefore\,\,\,$$ 100 $$\lambda $$ = 24800

$$ \Rightarrow \,\,\,\lambda $$ = 248

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