1

### 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  $\left| r \right| < 1$
2

### 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

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

${{{a_9}} \over {{a_5}}} = {{{a_1}{r^8}} \over {{a_1}{r^4}}} = {r^4} = {5^4}$
3

### 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)

${{a + 48d} \over {a + 28d}} = {{ - 18d + 48d} \over { - 18d + 28d}} = {3 \over 1}$
4

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

Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression ${{{x^m}{y^n}} \over {\left( {1 + {x^{2m}}} \right)\left( {1 + {y^{2n}}} \right)}}$ is :
A
${1 \over 2}$
B
${1 \over 4}$
C
${{m + n} \over {6mn}}$
D
1

## Explanation

${{{x^m}{y^n}} \over {\left( {1 + {x^{2m}}} \right)\left( {1 + {y^{2n}}} \right)}} = {1 \over {\left( {{x^m} + {1 \over {{x^m}}}} \right)\left( {{y^n} + {1 \over {{y^n}}}} \right)}} \le {1 \over 4}$

using AM $\ge$ GM