1

### JEE Main 2018 (Online) 15th April Evening Slot

Let    An = $\left( {{3 \over 4}} \right) - {\left( {{3 \over 4}} \right)^2} + {\left( {{3 \over 4}} \right)^3}$ $-$. . . . . + ($-$1)n-1 ${\left( {{3 \over 4}} \right)^n}$    and    Bn = 1 $-$ An.
Then, the least dd natural numbr p, so that Bn > An , for all n$\ge$ p, is :
A
9
B
7
C
11
D
5

## Explanation

An = $\left( {{3 \over 4}} \right) - {\left( {{3 \over 4}} \right)^2} + {\left( {{3 \over 4}} \right)^3} - .... + {\left( { - 1} \right)^{n - 1}}{\left( {{3 \over 4}} \right)^n}$

Which in a G.P. with a = ${{3 \over 4}}$, r = ${{{ - 3} \over 4}}$ and number of terms = n

$\therefore\,\,\,$ An = ${{{3 \over 4}\left( {1 - {{\left( {{{ - 3} \over 4}} \right)}^n}} \right)} \over {1 - \left( {{{ - 3} \over 4}} \right)}} = {{{3 \over 4} \times \left( {1 - {{\left( {{{ - 3} \over 4}} \right)}^n}} \right)} \over {{7 \over 4}}}$

$\Rightarrow $$\,\,\,An = {{3 \over 7}}$$\left[ {1 - {{\left( {{{ - 3} \over 4}} \right)}^n}} \right]$ $\,\,\,\,\,\,\,\,\,$ . . . . . . . . . .(1)

As, Bn = 1 $-$ An

For least odd natural number p, such that Bn > An

$\Rightarrow $$\,\,\, 1 - An > An \Rightarrow 1 > 2 \times An \Rightarrow An < {{1 \over 2}} From eqn. (1), we get {{3 \over 7}}$$ \times$ $\left[ {1 - {{\left( {{{ - 3} \over 4}} \right)}^n}} \right] < {1 \over 2}$ $\Rightarrow$ 1 $-$ ${\left( {{{ - 3} \over 4}} \right)^n} < {7 \over 6}$

$\Rightarrow $$\,\,\, 1 - {7 \over 6} < {\left( {{{ - 3} \over 4}} \right)^n} \Rightarrow {{ - 1} \over 6} < {\left( {{{ - 3} \over 4}} \right)^n} As n is odd, then {\left( {{{ - 3} \over 4}} \right)^n} = - {{{{3^n}} \over 4}} So {{ - 1} \over 6} < - {\left( {{3 \over 4}} \right)^n} \Rightarrow {1 \over 6} > {\left( {{3 \over 4}} \right)^n} log\left( {{1 \over 6}} \right) = n log\left( {{3 \over 4}} \right) \Rightarrow 6.228 < n Hence, n should be 7. 2 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 16th April Morning Slot The sum of the first 20 terms of the series 1 + {3 \over 2} + {7 \over 4} + {{15} \over 8} + {{31} \over {16}} + ..., is : A 38 + {1 \over {{2^{19}}}} B 38 + {1 \over {{2^{20}}}} C 39 + {1 \over {{2^{20}}}} D 39 + {1 \over {{2^{19}}}} ## Explanation 1 + {3 \over 2} + {7 \over 4} + {15 \over 8} + {31 \over 16} + . . . . = (2 - 1) + (2 - {1 \over 2} ) + (2 - {1 \over 4}) + (2 - {1 \over 8}) + . . . . .+ 20 terms = (2 + 2 + . . . . . 20 terms) - (1 + {1 \over 2} + {1 \over 4} + . . . . . 20 terms) = 2 \times 20 - \left( {{{1 - {{\left( {{1 \over 2}} \right)}^{20}}} \over {1 - {1 \over 2}}}} \right) = 40 - 2 + 2 {\left( {{1 \over 2}} \right)^{20}} = 38 + {1 \over {{2^{19}}}} 3 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 16th April Morning Slot Let {1 \over {{x_1}}},{1 \over {{x_2}}},...,{1 \over {{x_n}}}\,\, (xi \ne 0 for i = 1, 2, ..., n) be in A.P. such that x1=4 and x21 = 20. If n is the least positive integer for which {x_n} > 50, then \sum\limits_{i = 1}^n {\left( {{1 \over {{x_i}}}} \right)} is equal to : A {1 \over 8} B 3 C {{13} \over 8} D {{13} \over 4} ## Explanation \because$$\,\,\,$ ${1 \over {{x_1}}},{1 \over {{x_2}}},{1 \over {{x_3}}},.....,{1 \over {{x_n}}}$ 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}}}}$ = ${{25} \over 2}\left[ {\left( {2 \times {1 \over 4}} \right) + \left( {25 - 1} \right) \times \left( { - {1 \over {100}}} \right)} \right]$ = ${{13} \over 4}$
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### 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.