Joint Entrance Examination

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1

MCQ (Single Correct Answer)

The sum of the serier $${1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..............$$ up to $$\infty $$ is equal to

A

$$\log {\,_e}\left( {{4 \over e}} \right)\,\,$$

B

$$2\,\log {\,_e}2$$

C

$$\log {\,_e}2 - 1\,$$

D

$$\log {\,_e}2$$

$${1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..........\infty $$

$$\left| {{T_n}} \right| = {1 \over {n\left( {n + 1} \right)}} = \left( {{1 \over n} - {1 \over {n + 1}}} \right)$$

$$S = {T_1} - {T_2} + {T_3} - {T_4} + {T_5}.........\infty $$

$$ = \left( {{1 \over 1} - {1 \over 2}} \right) - \left( {{1 \over 2} - {1 \over 3}} \right) + \left( {{1 \over 3} - {1 \over 4}} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {{1 \over 4} - {1 \over 5}} \right).......$$

$$ = 1 - 2\left[ {{1 \over 2} - {1 \over 3} + {1 \over 4} - {1 \over 5}.........\infty } \right]$$

$$ = 1 - 2\left[ { - \log \left( {1 + 1} \right) + 1} \right]$$

$$ = 2\log 2 - 1 = \log \left( {{4 \over e}} \right)$$

$$\left| {{T_n}} \right| = {1 \over {n\left( {n + 1} \right)}} = \left( {{1 \over n} - {1 \over {n + 1}}} \right)$$

$$S = {T_1} - {T_2} + {T_3} - {T_4} + {T_5}.........\infty $$

$$ = \left( {{1 \over 1} - {1 \over 2}} \right) - \left( {{1 \over 2} - {1 \over 3}} \right) + \left( {{1 \over 3} - {1 \over 4}} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {{1 \over 4} - {1 \over 5}} \right).......$$

$$ = 1 - 2\left[ {{1 \over 2} - {1 \over 3} + {1 \over 4} - {1 \over 5}.........\infty } \right]$$

$$ = 1 - 2\left[ { - \log \left( {1 + 1} \right) + 1} \right]$$

$$ = 2\log 2 - 1 = \log \left( {{4 \over e}} \right)$$

2

MCQ (Single Correct Answer)

$${1^3} - \,\,{2^3} + {3^3} - {4^3} + ... + {9^3} = $$

A

425

B

- 425

C

475

D

- 475

$${1^3} - {2^3} + {3^3} - {4^3} + ...... + {9^3}$$

$$ = {1^3} + {2^3} + {3^3} + ...... + {9^3}$$

$$\,\,\,\,\,\,\,\,\,\, - 2\left( {{2^3} + {4^3} + {6^3} + {8^3}} \right)$$

$$ = {\left[ {{{9 \times 10} \over 2}} \right]^2} - {2.2^3}\left[ {{1^3} + {2^3} + {3^3} + {4^3}} \right]$$

$$ = {\left( {45} \right)^2} - 16.{\left[ {{{4 \times 5} \over 2}} \right]^2}$$

$$ = 2025 - 1600 = 425$$

$$ = {1^3} + {2^3} + {3^3} + ...... + {9^3}$$

$$\,\,\,\,\,\,\,\,\,\, - 2\left( {{2^3} + {4^3} + {6^3} + {8^3}} \right)$$

$$ = {\left[ {{{9 \times 10} \over 2}} \right]^2} - {2.2^3}\left[ {{1^3} + {2^3} + {3^3} + {4^3}} \right]$$

$$ = {\left( {45} \right)^2} - 16.{\left[ {{{4 \times 5} \over 2}} \right]^2}$$

$$ = 2025 - 1600 = 425$$

3

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 $$100\left( {1 - r} \right)\left( {1 + r} \right)$$

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

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

$$ \Rightarrow 5r = 3$$

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

4

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

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

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

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

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

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

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