1

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

The sum of the following series

$1 + 6 + {{9\left( {{1^2} + {2^2} + {3^2}} \right)} \over 7} + {{12\left( {{1^2} + {2^2} + {3^2} + {4^2}} \right)} \over 9}$

$+ {{15\left( {{1^2} + {2^2} + ... + {5^2}} \right)} \over {11}} + .....$ up to 15 terms, is :
A
7520
B
7510
C
7830
D
7820

## Explanation

$1 + 6 + {{9\left( {{1^2} + {2^2} + {3^2}} \right)} \over 7} + {{12\left( {{1^2} + {2^2} + {3^2} + {4^2}} \right)} \over 9} + {{15\left( {{1^2} + {2^2} + ... + {5^2}} \right)} \over {11}} + .....\,15$

$= {{3\left( {{1^2}} \right)} \over 3} + {{6\left( {{1^2} + {2^2}} \right)} \over 5} + {{9\left( {{1^2} + {2^2} + {3^2}} \right)} \over 7} + {{12} \over 9}\left( {{1^2} + {2^2} + {3^2} + {4^2}} \right) + ......$

${T_r} = {{3r} \over {2r + 1}}\left( {{1^2} + {2^2} + .... + {r^2}} \right)$

${T_r} = {{3r} \over {2r + 1}}{{r\left( {r + 1} \right)\left( {2r + 1} \right)} \over 6} = {1 \over 2}{r^2}\left( {r + 1} \right)$

Sum of $n$ terms $= \sum\limits_{r = 1}^n {{T_r}} = {1 \over 2}\sum\limits_{r = 1}^n {\left( {{r^3} + {r^2}} \right)}$

$= {1 \over 2}\left[ {{{{n^2}{{\left( {n + 1} \right)}^2}} \over 4} + {{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6}} \right]$

Sum upto 15 terms $\Rightarrow$ then put $n$ = 15

$= {1 \over 2}\left( {{{{{\left( {15 \times 16} \right)}^2}} \over 4} + {{15 \times 16 \times 31} \over 6}} \right) = 7820$
2

### 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

${a \over c} = {{A + 6d} \over {A + 12d}} = {{ - 8d} \over { - 2d}} = 4$
3

### 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
4

### 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