Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is

A

5

B

3/5

C

8/5

D

1/5

Let $$a=$$ first team of $$G.P.$$ and $$r=$$ common ratio of $$G.P.;$$

Then $$G.P.$$ is $$a,$$ $$ar,$$ $$a{r^2}$$

Given $${S_\infty } = 20 \Rightarrow {a \over {1 - r}} = 20$$

$$ \Rightarrow a = 20\left( {1 - r} \right)....\left( i \right)$$

Also $${a^2} + {a^2}{r^2} + {a^2}{r^4} + ...$$ to $$\infty = 100$$

$$ \Rightarrow {{{a^2}} \over {1 - {r^2}}} = 100$$

$$ \Rightarrow {a^2} = 100\left( {1 - r} \right)\left( {1 + r} \right)....\left( {ii} \right)$$

From $$(i),$$ $${a^2} = 400{\left( {1 - r} \right)^2};$$

From $$(ii),$$ we get $$100\left( {1 - r} \right)\left( {1 + r} \right)$$

$$\,\,\,\,\,\,\,\,\,\, = 400{\left( {1 - r} \right)^2}$$

$$ \Rightarrow 1 + r = 4 - 4r$$

$$ \Rightarrow 5r = 3$$

$$ \Rightarrow r = 3/5.$$

Then $$G.P.$$ is $$a,$$ $$ar,$$ $$a{r^2}$$

Given $${S_\infty } = 20 \Rightarrow {a \over {1 - r}} = 20$$

$$ \Rightarrow a = 20\left( {1 - r} \right)....\left( i \right)$$

Also $${a^2} + {a^2}{r^2} + {a^2}{r^4} + ...$$ to $$\infty = 100$$

$$ \Rightarrow {{{a^2}} \over {1 - {r^2}}} = 100$$

$$ \Rightarrow {a^2} = 100\left( {1 - r} \right)\left( {1 + r} \right)....\left( {ii} \right)$$

From $$(i),$$ $${a^2} = 400{\left( {1 - r} \right)^2};$$

From $$(ii),$$ we get $$100\left( {1 - r} \right)\left( {1 + r} \right)$$

$$\,\,\,\,\,\,\,\,\,\, = 400{\left( {1 - r} \right)^2}$$

$$ \Rightarrow 1 + r = 4 - 4r$$

$$ \Rightarrow 5r = 3$$

$$ \Rightarrow r = 3/5.$$

2

MCQ (Single Correct Answer)

Fifth term of a GP is 2, then the product of its 9 terms is

A

256

B

512

C

1024

D

none of these

$$a{r^4} = 2$$

$$a \times ar \times a{r^2} \times a{r^3} \times a{r^4} \times a{r^5} \times a{r^6} \times a{r^7} \times a{r^8}$$

$$ = {a^9}{r^{36}} = {\left( {a{r^4}} \right)^9} = {2^9} = 512$$

$$a \times ar \times a{r^2} \times a{r^3} \times a{r^4} \times a{r^5} \times a{r^6} \times a{r^7} \times a{r^8}$$

$$ = {a^9}{r^{36}} = {\left( {a{r^4}} \right)^9} = {2^9} = 512$$

3

MCQ (Single Correct Answer)

The value of $$\,{2^{1/4}}.\,\,{4^{1/8}}.\,{8^{1/16}}...\infty $$ is

A

1

B

2

C

3/2

D

4

The product is $$p = {2^{1/4}}{.2^{2/8}}{.2^{3/16}}........$$

$$ = {2^{1/4 + 2/8 + 3/16 + .......\infty }}$$

Now let

$$S = {1 \over 4} + {2 \over 8} + {3 \over {16}} + .......\infty \,\,\,\,........\left( 1 \right)$$

$${1 \over 2}S = {1 \over 8} + {2 \over {16}} + .......\infty \,\,\,\,........\left( 2 \right)$$

Subtracting $$(2)$$ from $$(1)$$

$$ \Rightarrow {1 \over 2}S = {1 \over 4} + {1 \over 8} + {1 \over {16}} + .......\infty $$

or $${1 \over 2}S = {{1/4} \over {1 - 1/2}} = {1 \over 2} \Rightarrow S = 1$$

$$\therefore$$ $$P = {2^S} = 2$$

$$ = {2^{1/4 + 2/8 + 3/16 + .......\infty }}$$

Now let

$$S = {1 \over 4} + {2 \over 8} + {3 \over {16}} + .......\infty \,\,\,\,........\left( 1 \right)$$

$${1 \over 2}S = {1 \over 8} + {2 \over {16}} + .......\infty \,\,\,\,........\left( 2 \right)$$

Subtracting $$(2)$$ from $$(1)$$

$$ \Rightarrow {1 \over 2}S = {1 \over 4} + {1 \over 8} + {1 \over {16}} + .......\infty $$

or $${1 \over 2}S = {{1/4} \over {1 - 1/2}} = {1 \over 2} \Rightarrow S = 1$$

$$\therefore$$ $$P = {2^S} = 2$$

4

MCQ (Single Correct Answer)

l, m, n are the $${p^{th}}$$, $${q^{th}}$$ and $${r^{th}}$$ term of a G.P all positive, $$then\,\left| {\matrix{
{\log \,l} & p & 1 \cr
{\log \,m} & q & 1 \cr
{\log \,n} & r & 1 \cr
} } \right|\,equals$$

A

- 1

B

2

C

1

D

0

$$l = A{R^{p - 1}}$$

$$ \Rightarrow \log 1 = \log A + \left( {p - 1} \right)\log R$$

$$m = A{R^{q - 1}}$$

$$ \Rightarrow \log m = \log A + \left( {q - 1} \right)\log R$$

$$n = A{R^{r - 1}}$$

$$ \Rightarrow \log n = \log A + \left( {r - 1} \right)\log R$$

Now, $$\left| {\matrix{ {\log l} & p & 1 \cr {\log m} & q & 1 \cr {\log n} & r & 1 \cr } } \right|$$

$$ = \left| {\matrix{ {\log A + \left( {p - 1} \right)\log R} & p & 1 \cr {\log A + \left( {q - 1} \right)\log R} & q & 1 \cr {\log A + \left( {r - 1} \right)\log R} & r & 1 \cr } } \right|$$

Operating $${C_1} - \left( {\log R} \right){C_2} + \left( {\log R - \log A} \right){C_3}$$

$$ = \left| {\matrix{ 0 & p & 1 \cr 0 & q & 1 \cr 0 & r & 1 \cr } } \right| = 0$$

$$ \Rightarrow \log 1 = \log A + \left( {p - 1} \right)\log R$$

$$m = A{R^{q - 1}}$$

$$ \Rightarrow \log m = \log A + \left( {q - 1} \right)\log R$$

$$n = A{R^{r - 1}}$$

$$ \Rightarrow \log n = \log A + \left( {r - 1} \right)\log R$$

Now, $$\left| {\matrix{ {\log l} & p & 1 \cr {\log m} & q & 1 \cr {\log n} & r & 1 \cr } } \right|$$

$$ = \left| {\matrix{ {\log A + \left( {p - 1} \right)\log R} & p & 1 \cr {\log A + \left( {q - 1} \right)\log R} & q & 1 \cr {\log A + \left( {r - 1} \right)\log R} & r & 1 \cr } } \right|$$

Operating $${C_1} - \left( {\log R} \right){C_2} + \left( {\log R - \log A} \right){C_3}$$

$$ = \left| {\matrix{ 0 & p & 1 \cr 0 & q & 1 \cr 0 & r & 1 \cr } } \right| = 0$$

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