1

### JEE Main 2019 (Online) 10th January Evening Slot

Let a1, a2, a3, ..... a10 be in G.P. with ai > 0 for i = 1, 2, ….., 10 and S be the set of pairs (r, k), r, k $\in$ N (the set of natural numbers) for which

$\left| {\matrix{ {{{\log }_e}\,{a_1}^r{a_2}^k} & {{{\log }_e}\,{a_2}^r{a_3}^k} & {{{\log }_e}\,{a_3}^r{a_4}^k} \cr {{{\log }_e}\,{a_4}^r{a_5}^k} & {{{\log }_e}\,{a_5}^r{a_6}^k} & {{{\log }_e}\,{a_6}^r{a_7}^k} \cr {{{\log }_e}\,{a_7}^r{a_8}^k} & {{{\log }_e}\,{a_8}^r{a_9}^k} & {{{\log }_e}\,{a_9}^r{a_{10}}^k} \cr } } \right|$ $=$ 0.

Then the number of elements in S, is -
A
10
B
4
C
2
D
infinitely many

## Explanation

Apply

C3 $\to$ C3 $-$ C2

C2 $\to$ C2 $-$ C1

We get D = 0
2

### JEE Main 2019 (Online) 11th January Morning Slot

The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is ${{27} \over {19}}$.Then the common ratio of this series is :
A
${4 \over 9}$
B
${1 \over 3}$
C
${2 \over 3}$
D
${2 \over 9}$

## Explanation

${a \over {1 - r}} = 3\,\,\,\,\,\,\,.....(1)$

${{{a^3}} \over {1 - {r^3}}} = {{27} \over {19}} \Rightarrow {{27{{\left( {1 - r} \right)}^3}} \over {1 - {r^3}}} = {{27} \over {19}}$

$\Rightarrow 6{r^2} - 13r + 6 = 0$

$\Rightarrow r = {2 \over 3}\,\,$

as  $\left| r \right| < 1$
3

### JEE Main 2019 (Online) 11th January Morning Slot

Let a1, a2, . . . . . ., a10 be a G.P.    If ${{{a_3}} \over {{a_1}}} = 25,$ then ${{{a_9}} \over {{a_5}}}$ equals
A
53
B
2(52)
C
4(52)
D
54

## Explanation

a1, a2, . . . . ., a10 are in G.P.,

Let the common ratio be r

${{{a_3}} \over {{a_1}}} = 25 \Rightarrow {{{a_1}{r^2}} \over {{a_1}}} = 25 \Rightarrow {r^2} = 25$

${{{a_9}} \over {{a_5}}} = {{{a_1}{r^8}} \over {{a_1}{r^4}}} = {r^4} = {5^4}$
4

### JEE Main 2019 (Online) 11th January Evening Slot

If 19th term of a non-zero A.P. is zero, then its (49th term) : (29th term) is :
A
2 : 1
B
4 : 1
C
1 : 3
D
3 : 1

## Explanation

a + 18d = 0        . . . . .(1)

${{a + 48d} \over {a + 28d}} = {{ - 18d + 48d} \over { - 18d + 28d}} = {3 \over 1}$