1

### JEE Main 2019 (Online) 9th January Evening Slot

Let a, b and c be the 7th, 11th and 13th terms respectively of a non-constant A.P. If these are also three consecutive terms of a G.P., then ${a \over c}$ equal to :
A
2
B
${1 \over 2}$
C
${7 \over 13}$
D
4

## Explanation

T7 = A + 6d = a; T11 = A + 10d = b; T13 = A + 12d = c

Now a, b, c are in G.P.

$\therefore$  b2 = ac

$\Rightarrow$  (A + 10d)2 = (A + 6d) (A + 12d)

$\Rightarrow$  A2 + 100d2 + 20Ad = A2 + 18Ad + 72d2

$\Rightarrow$  A + 14d = 0, A = $-$ 14d

${a \over c} = {{A + 6d} \over {A + 12d}} = {{ - 8d} \over { - 2d}} = 4$
2

### 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
3

### 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
4

### 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

### EXAM MAP

#### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE ME GATE PI GATE EE GATE CE GATE IN

NEET