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1

### JEE Main 2014 (Offline)

Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is :
A
$$2 - \sqrt 3$$
B
$$2 + \sqrt 3$$
C
$$\sqrt 2 + \sqrt 3$$
D
$$3 + \sqrt 2$$

## Explanation

Let $$a,ar,a{r^2}$$ are in $$G.P.$$

According to the question

$$a,2ar,a{r^2}$$ are in $$A.P.$$

$$\Rightarrow 2 \times 2ar = a + a{r^2}$$

$$\Rightarrow 4r = 1 + {r^2}$$

$$\Rightarrow {r^2} - 4r + 1 = 0$$

$$r = {{4 \pm \sqrt {16 - 4} } \over 2} = 2 \pm \sqrt 3$$

Since $$r > 1$$

$$\therefore$$ $$\pi = 2 - \sqrt 3$$ is rejected

Hence, $$r = 2 + \sqrt 3$$
2

### JEE Main 2013 (Offline)

The sum of first 20 terms of the sequence 0.7, 0.77, 0.777,........,is
A
$${7 \over {81}}\left( {179 - {{10}^{ - 20}}} \right)$$
B
$$\,{7 \over 9}\left( {99 - {{10}^{ - 20}}} \right)$$
C
$${7 \over {81}}\left( {179 + {{10}^{ - 20}}} \right)$$
D
$${7 \over 9}\left( {99 + {{10}^{ - 20}}} \right)$$

## Explanation

Given sequence can be written as

$${7 \over {10}} + {{77} \over {100}} + {{777} \over {{{10}^3}}} + ..... +$$ up to $$20$$ terms

$$= 7\left[ {{1 \over {10}} + {{11} \over {100}} + {{111} \over {{{10}^3}}} + ...... + } \right.\,\,$$ up to $$20$$ terms ]

Multiply and divide by $$9$$

$$= {7 \over 9}\left[ {{9 \over {10}} + {{99} \over {100}} + {{999} \over {1000}} + ......} \right.\,\,$$ $$+$$ up to $$20$$ terms ]

$$= {7 \over 9}\left[ {\left( {1 - {1 \over {10}}} \right)} \right. + \left( {1 - {1 \over {{{10}^2}}}} \right) + \left( {1 - {1 \over {{{10}^3}}}} \right) + ......$$ $$+$$ up to $$20$$ terms ]

$$= {7 \over 9}\left[ {20 - {{{1 \over {10}}\left( {1 - {{\left( {{1 \over {10}}} \right)}^{20}}} \right)} \over {1 - {1 \over {10}}}}} \right]$$

$$= {7 \over 9}\left[ {{{179} \over 9} + {1 \over 9}{{\left( {{1 \over {10}}} \right)}^{20}}} \right]$$

$$= {7 \over {81}}\left[ {179 + {{\left( {10} \right)}^{ - 20}}} \right]$$
3

### AIEEE 2012

Statement-1: The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) +.....+ (361 + 380 + 400) is 8000.

Statement-2: $$\sum\limits_{k = 1}^n {\left( {{k^3} - {{(k - 1)}^3}} \right)} = {n^3}$$, for any natural number n.

A
Statement-1 is false, Statement-2 is true.
B
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
C
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
D
Statement-1 is true, Statement-2 is false.

## Explanation

$$n$$th term of the given series

$$= {T_n} = {\left( {n - 1} \right)^2} + \left( {n - 1} \right)n + {n^2}$$

$$= {{\left( {{{\left( {n - 1} \right)}^3} - {n^3}} \right)} \over {\left( {n - 1} \right) - n}}$$

$$= {n^3} - {\left( {n - 1} \right)^3}$$

$$\Rightarrow {S_n} = \sum\limits_{k = 1}^n {\left[ {{k^3} - {{\left( {k - 1} \right)}^3}} \right]}$$

$$\Rightarrow 8000 = {n^3}$$

$$\Rightarrow n = 20\,\,$$ which is a natural number.

Now, put $$n = 1,2,3,.....20$$

$${T_1} = {1^3} - {0^3}$$

$${T_2} = {2^3} - {1^3}$$

.

.

.

$${T_{20}} = {20^3} - {19^3}$$

Now, $${T_1} + {T_2} + ..... + {T_{20}} = {S_{20}}$$

$$\Rightarrow {S_{20}} = {20^3} - {0^3} = 8000$$

Hence, both the given statements are true and statement $$2$$ supports statement $$1.$$
4

### AIEEE 2011

A man saves ₹ 200 in each of the first three months of his service. In each of the subsequent months his saving increases by ₹ 40 more than the saving of immediately previous month. His total saving from the start of service will be ₹ 11040 after
A
19 months
B
20 months
C
21 months
D
18 months

## Explanation

Let required number of months $$=n$$

$$\therefore$$ $$200 \times 3 + \left( {240 + 280 + 320 + ...} \right.$$

$$\left. {\,\,\,\,\,\,\,\,\,\,\,\, + {{\left( {n - 3} \right)}^{th}}\,term} \right) = 11040$$

$$\Rightarrow {{n - 3} \over 2}\left[ {2 \times 240 + \left( {n - 4} \right) \times 40} \right]$$

$$\,\,\,\,\,\,\,\,\,\,\,\, = 11040 - 600$$

$$\Rightarrow \left( {n - 3} \right)\left[ {240 + 20n - 80} \right] = 10440$$

$$\Rightarrow \left( {n - 3} \right)\left( {20n + 160} \right) = 10440$$

$$\Rightarrow \left( {n - 3} \right)\left( {n + 8} \right) = 522$$

$$\Rightarrow {n^2} + 5n - 546 = 0$$

$$\Rightarrow \left( {n + 26} \right)\left( {n - 21} \right) = 0$$

$$\therefore$$ $$n = 21$$

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