1

JEE Main 2019 (Online) 9th January Morning Slot

Let ${a_1},{a_2},.......,{a_{30}}$ be an A.P.,

$S = \sum\limits_{i = 1}^{30} {{a_i}}$ and $T = \sum\limits_{i = 1}^{15} {{a_{\left( {2i - 1} \right)}}}$.

If $a_5$ = 27 and S - 2T = 75, then $a_{10}$ is equal to :
A
47
B
42
C
52
D
57

Explanation

Let the common difference = d

S = $\sum\limits_{i = 1}^{30} {{a_i}}$

= $a$1 + $a$2 + . . . . . + $a$30

$\therefore$  S = ${{30} \over 2}\left[ {{a_1} + {a_{30}}} \right]$

= 15 [$a$1 + $a$1 + 29d]

= 15 (2$a$1 + 29d)

T = $\sum\limits_{i = 1}^{15} {{a_{\left( {2i - 1} \right)}}}$

= $a$1 + $a$3 + . . . . . . + $a$29

= ${{15} \over 2}\left[ {a{}_1 + {a_{29}}} \right]$

= ${{15} \over 2}\left[ {a{}_1 + {a_1} + 28d} \right]$

= ${{15} \over 2}\left[ {2a{}_1 + 28d} \right]$

= 15 ($a$1 + 14d)

Given,

S $-$ 2T = 75

$\Rightarrow$  15(2$a$1 + 29d) $-$ 2 $\times$ 15 ($a$1 + 14d) = 75

$\Rightarrow$  30$a$1 + 15 $\times$ 29d $-$ 30 $a$1 $-$ 420d = 75

$\Rightarrow$  435d $-$ 420d = 75

$\Rightarrow$  15d = 75

$\Rightarrow$  d = 5

Given that,

$a$5 = 27

$\Rightarrow$  $a$1 + 4d = 27

$\Rightarrow$  $a$1 + 20 = 27

$\Rightarrow$  $a$1 = 7

$\therefore$  $a$10 = $a$1 + 9d

= 7 + 45

= 52
2

JEE Main 2019 (Online) 10th January Morning Slot

The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is -
A
1356
B
1256
C
1365
D
1465

Explanation

$\sum\limits_{r = 2}^{13} {(7r + 2) = 7.{{2 + 13} \over 2}} \times 6 + 2 \times 12$

= 7 $\times$90 + 24 = 654

$\sum\limits_{r = 1}^{13} {(7r + 5) = 7\left( {{{1 + 13} \over 2}} \right)} \times 13 + 5 \times 13 = 702$

Total = 654 + 702 = 1356
3

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
4

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  $\left| r \right| < 1$