1

### JEE Main 2018 (Online) 16th April Morning Slot

The sum of the first 20 terms of the series

$1 + {3 \over 2} + {7 \over 4} + {{15} \over 8} + {{31} \over {16}} + ...,$ is :
A
$38 + {1 \over {{2^{19}}}}$
B
$38 + {1 \over {{2^{20}}}}$
C
$39 + {1 \over {{2^{20}}}}$
D
$39 + {1 \over {{2^{19}}}}$

## Explanation

1 + ${3 \over 2}$ + ${7 \over 4}$ + ${15 \over 8}$ + ${31 \over 16}$ + . . . .

=  (2 $-$ 1) + (2 $-$ ${1 \over 2}$ ) + (2 $-$ ${1 \over 4}$) + (2 $-$ ${1 \over 8}$) + . . . . .+ 20 terms

=   (2 + 2 + . . . . . 20 terms) $-$ (1 + ${1 \over 2}$ + ${1 \over 4}$ + . . . . . 20 terms)

=   2 $\times$ 20 $-$ $\left( {{{1 - {{\left( {{1 \over 2}} \right)}^{20}}} \over {1 - {1 \over 2}}}} \right)$

=   40 $-$ 2 + 2 ${\left( {{1 \over 2}} \right)^{20}}$

=  38 + ${1 \over {{2^{19}}}}$
2

### JEE Main 2018 (Online) 16th April Morning Slot

Let ${1 \over {{x_1}}},{1 \over {{x_2}}},...,{1 \over {{x_n}}}\,\,$ (xi $\ne$ 0 for i = 1, 2, ..., n) be in A.P. such that x1=4 and x21 = 20. If n is the least positive integer for which ${x_n} > 50,$ then $\sum\limits_{i = 1}^n {\left( {{1 \over {{x_i}}}} \right)}$ is equal to :
A
${1 \over 8}$
B
3
C
${{13} \over 8}$
D
${{13} \over 4}$

## Explanation

$\because $$\,\,\, {1 \over {{x_1}}},{1 \over {{x_2}}},{1 \over {{x_3}}},.....,{1 \over {{x_n}}} are in A.P. x1 = 4 and x21 = 20 Let 'd' be the common difference of this A.P. \therefore\,\,\, its 21st term = {1 \over {{x_{21}}}} = {1 \over {{x_1}}} + \left[ {\left( {21 - 1} \right) \times d} \right] \Rightarrow$$\,\,\,$ d = ${1 \over {20}}$ $\times$ $\left( {{1 \over {20}} - {1 \over 4}} \right)$ $\Rightarrow$ d = $-$ ${1 \over {100}}$

Also xn > 50(given).

$\therefore\,\,\,$ ${1 \over {{x_n}}} = {1 \over {{x_1}}} + \left[ {\left( {n - 1} \right) \times d} \right]$

$\Rightarrow $$\,\,\, xn = {{{x_1}} \over {1 + \left( {n - 1} \right) \times d \times {x_1}}} \therefore\,\,\, {{{x_1}} \over {1 + \left( {n - 1} \right) \times d \times {x_1}}} > 50 \Rightarrow$$\,\,\,$ ${4 \over {1 + \left( {n - 1} \right) \times \left( { - {1 \over {100}}} \right) \times 4}} > 50$

$\Rightarrow $$\,\,\, 1 + (n - 1) \times (- {1 \over {100}}) \times 4 < {4 \over {50}} \Rightarrow$$\,\,\,$ $-$ ${1 \over {100}}$(n $-$ 1) < $-$ ${{23} \over {100}}$

$\Rightarrow $$\,\,\, n - > 23 \Rightarrow n > 24 Therefore\,\,\, n = 25. \Rightarrow$$\,\,\,$$\sum\limits_{i = 1}^{25} {{1 \over {{x_i}}}}$ = ${{25} \over 2}\left[ {\left( {2 \times {1 \over 4}} \right) + \left( {25 - 1} \right) \times \left( { - {1 \over {100}}} \right)} \right]$ = ${{13} \over 4}$
3

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

If a, b, c be three distinct real numbers in G.P. and a + b + c = xb , then x cannot be
A
2
B
-3
C
4
D
-2

## Explanation

a, b, c are in G.P.

So, b = ar

and c = ar2

given   a + b + c = xb

$\Rightarrow$  a + br + ar2 = x(ar)

$\Rightarrow$  1 + r + r2 = xr

$\Rightarrow$  x = 1 + r + ${1 \over r}$

let sum of r + ${1 \over r}$ = M

$\therefore$  r2 + 1 = Mr

$\Rightarrow$  r2 $-$ Mr + 1 = 0

real solution when discriminant is $\ge$ 0

$\therefore$  b2 $-$ 4ac $\ge$ 0

M2 $-$ 4.1.1 $\ge$ 0

$\Rightarrow$  M2 $\ge$ 4

M $\ge$ 2 or M $\le$ $-$ 2

$\therefore$  M $\in$ ($-$ $\propto$, $-$ 2] $\cup$ [2, $\propto$)

As   x = 1 + r + ${1 \over r}$

= 1 + M

$\therefore$  x $\in$ ($-$ $\propto$, $-$ 1] $\cup$ [3, $\propto$)

$\therefore$  x can't be 0, 1, 2.
4

### 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

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