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

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

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

2

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

3

Let $${a_1}$$, $${a_2}$$, $${a_3}$$, ......... ,$${a_{49}}$$ be in A.P. such that

$$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416$$ and $${a_9} + {a_{43}} = 66$$.

$$a_1^2 + a_2^2 + ....... + a_{17}^2 = 140m$$, then m is equal to

$$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416$$ and $${a_9} + {a_{43}} = 66$$.

$$a_1^2 + a_2^2 + ....... + a_{17}^2 = 140m$$, then m is equal to

A

33

B

66

C

68

D

34

a_{1}, a_{2}, a_{3} . . . a_{43} are in AP

So, a_{2} = a_{1} + d

a_{3} = a_{1} + 2d

.

.

.

a_{49} =a_{1} + 48d

Now given, $${a_9} + {a_{43}} = 66$$

$$ \Rightarrow \,\,\,\,$$ a_{1} + 8d + a_{1} + 42d = 66

$$ \Rightarrow \,\,\,\,$$ 2a_{1} + 50d = 66

$$ \Rightarrow \,\,\,\,$$ a_{1} + 25d = 33 . . . . . (1)

$$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} $$ = 416

$$ \Rightarrow \,\,\,\,$$ a_{1} + a_{5} + a_{9} + a_{13} +. . . . . 13 items = 416

$$ \Rightarrow \,\,\,\,$$ a_{1} + a_{1} + 4d + a_{1} + 8d + . . . . a_{1} + 48d = 416

$$ \Rightarrow \,\,\,\,$$ 13a_{1} + 4d +8d + 12d + . . . . . 48d = 416

$$ \Rightarrow \,\,\,\,$$ 13a_{1} + 4 (1+ 2 + 3 + . . . + 12) d = 416

$$ \Rightarrow \,\,\,\,13\,\,a{}_1 + \,4\,\, \times \,{{12 \times 13} \over 2} \times $$d = 416

$$ \Rightarrow \,\,\,\,$$ 13a_{1} + 24 $$ \times$$ 13d = 416

$$ \Rightarrow \,\,\,\,$$ a_{1} + 24 d =32 . . . .(2)

Solving (1) and (2) we get,

d = 1

and $${a_1} = 8$$

$$\therefore\,\,\,$$ a_{1} = 8

a_{2} = 8 + 1 = 9

a_{3} = 8 + 2 = 10

.

.

.

a_{17} = 8 + 16 = 24

Now, $$a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2\,\, = \,\,140m$$

$$ \Rightarrow \,\,\,\,$$ $$a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2 = 140\,m$$

$$ \Rightarrow \,\,\,\,\,{8^2}\, + \,\,{9^2}\, + \,{10^2} + ......{(24)^2} = 140\,m$$

We can write above series like this,

$$ \Rightarrow \,\,\,\,\,$$ (1^{2} +2^{2} + . . . . +24^{2}) $$-$$ (1^{2} + 2^{2} + . . . . .+ 7^{2}) = 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

So, a

a

.

.

.

a

Now given, $${a_9} + {a_{43}} = 66$$

$$ \Rightarrow \,\,\,\,$$ a

$$ \Rightarrow \,\,\,\,$$ 2a

$$ \Rightarrow \,\,\,\,$$ a

$$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} $$ = 416

$$ \Rightarrow \,\,\,\,$$ a

$$ \Rightarrow \,\,\,\,$$ a

$$ \Rightarrow \,\,\,\,$$ 13a

$$ \Rightarrow \,\,\,\,$$ 13a

$$ \Rightarrow \,\,\,\,13\,\,a{}_1 + \,4\,\, \times \,{{12 \times 13} \over 2} \times $$d = 416

$$ \Rightarrow \,\,\,\,$$ 13a

$$ \Rightarrow \,\,\,\,$$ a

Solving (1) and (2) we get,

d = 1

and $${a_1} = 8$$

$$\therefore\,\,\,$$ a

a

a

.

.

.

a

Now, $$a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2\,\, = \,\,140m$$

$$ \Rightarrow \,\,\,\,$$ $$a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2 = 140\,m$$

$$ \Rightarrow \,\,\,\,\,{8^2}\, + \,\,{9^2}\, + \,{10^2} + ......{(24)^2} = 140\,m$$

We can write above series like this,

$$ \Rightarrow \,\,\,\,\,$$ (1

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

$$ \Rightarrow \,\,\,\,\,$$ 490 $$-$$ 140 = 140 m

$$ \Rightarrow \,\,\,\,\,$$4760 = 140 m

$$ \Rightarrow \,\,\,\,\,$$ m = 34

4

If x_{1}, x_{2}, . . ., x_{n} and $${1 \over {{h_1}}}$$, $${1 \over {{h_2}}}$$, . . . , $${1 \over {{h_n}}}$$ are two A.P..s such that x_{3} = h_{2} = 8 and x_{8} = h_{7} = 20, then x_{5}.h_{10} equals :

A

2560

B

2650

C

3200

D

1600

Assume d_{1} is the common difference of A.P x_{1},x_{2} ..... x_{n}

Given x_{3} = 8 and x_{8} = 20

$$ \therefore $$ x_{1} + 2d_{1} = 8 ..... **(i)**

and x_{1} + 7d_{1} = 20 .....** (ii)**

Solving**(i)** and **(ii)** we get x_{1} = $$16 \over {15}$$ and d_{1} = $$12 \over {5}$$

Now let $$1 \over d_2$$ is the common difference of A.P $$1 \over h_1$$, $$1 \over h_2$$ ..... $$1 \over h_n$$

Given that,

h_{2} = 8 and h_{7} = 20

$$ \therefore $$ $$1 \over h_2$$ = $$1 \over 8$$

$$ \Rightarrow $$ $$1 \over h_1$$ + $$1 \over d_2$$ = $$1 \over 8$$ ....**(iii)**

and $$1 \over h_7$$ = $$1 \over 20$$

$$ \Rightarrow $$ $$1 \over h_1$$ + $$6 \over d_2$$ = $$1 \over 20$$ ...**(iv)**

Solving**(iii)** and **(iv)** we get

$$1 \over h_1$$ = $$28 \over 200$$ and $$1 \over d_2$$ = $$- {3 \over 200}$$

So, x_{5} = x_{1} + 4d_{1}

= $$16 \over 5$$ + $$48 \over 5$$= $$64 \over 5$$ and

$$1 \over h_{10}$$ = $$1 \over h_1$$ + $$9 \over d_2$$

= $$28 \over 200$$ - $$27 \over 200$$ = $$1\over 200$$

$$ \therefore $$ x_{5} $$\times$$ h_{10} = $${64 \over 5} \times 200$$ = 2560

Given x

$$ \therefore $$ x

and x

Solving

Now let $$1 \over d_2$$ is the common difference of A.P $$1 \over h_1$$, $$1 \over h_2$$ ..... $$1 \over h_n$$

Given that,

h

$$ \therefore $$ $$1 \over h_2$$ = $$1 \over 8$$

$$ \Rightarrow $$ $$1 \over h_1$$ + $$1 \over d_2$$ = $$1 \over 8$$ ....

and $$1 \over h_7$$ = $$1 \over 20$$

$$ \Rightarrow $$ $$1 \over h_1$$ + $$6 \over d_2$$ = $$1 \over 20$$ ...

Solving

$$1 \over h_1$$ = $$28 \over 200$$ and $$1 \over d_2$$ = $$- {3 \over 200}$$

So, x

= $$16 \over 5$$ + $$48 \over 5$$= $$64 \over 5$$ and

$$1 \over h_{10}$$ = $$1 \over h_1$$ + $$9 \over d_2$$

= $$28 \over 200$$ - $$27 \over 200$$ = $$1\over 200$$

$$ \therefore $$ x

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