1

### 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}}}}$ = ${{25} \over 2}\left[ {\left( {2 \times {1 \over 4}} \right) + \left( {25 - 1} \right) \times \left( { - {1 \over {100}}} \right)} \right]$ = ${{13} \over 4}$
2

### 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.
3

### 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

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

### 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

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

NEET