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1

JEE Advanced 2016 Paper 2 Offline

MCQ (Single Correct Answer)

Let bi > 1 for I = 1, 2, ......, 101. Suppose logeb1, logeb2, ......., logeb101 are in Arithmetic Progression (A.P.) with the common difference loge2. Suppose a1, a2, ......, a101 are in A.P. such that a1 = b1 and a51 = b51. If t = b1 + b2 + .... + b51 and s = a1 + a2 + ..... + a51, then

A
s > t and a101 > b101
B
s > t and a101 < b101
C
s < t and a101 > b101
D
s < t and a101 < b101

Explanation

If logb1, logb2, ......, logb101 are in A.P. with common difference loge2, then b1, b2, ......, b101 are in G.P., with common ratio 2.

$$\therefore$$ b1 = 20b1

b2 = 21b1

b3 = 22b1

$$\matrix{ \vdots & \vdots & \vdots \cr } $$

b101 = 2100b1 ..... (i)

Also, a1, a2, ......, a101 are in A.P.

Given, a1 = b1 and a51 = b51

$$\Rightarrow$$ a1 = b1 and a51 = b51

$$\Rightarrow$$ a1 + 50D = 250 b1

$$\Rightarrow$$ a1 + 50D = 250 a1 [$$\because$$ a1 = b1] ...... (ii)

Now, t = b1 + b2 + ..... + b51

$$ \Rightarrow t = {b_1}{{({2^{51}} - 1)} \over {2 - 1}}$$ ..... (iii)

and s = a1 + a2 + .... + a51

$$ = {{51} \over 2}(2{a_1} + 50D)$$ ...... (iv)

$$\therefore$$ t = a1(251 $$-$$ 1) [$$\because$$ a1 = b1]

or t = 251 a1 $$-$$ a1 < 251 a1 ...... (v)

and $$s = {{51} \over 2}[{a_1} + ({a_1} + 50D)]$$ [from Eq. (ii)]

$$ = {{51} \over 2}[{a_1} + {2^{50}}{a_1}] = {{51} \over 2}{a_1} + {{51} \over 2}{2^{50}}{a_1}$$

$$\therefore$$ s > 251 a1 ...... (vi)

From Eqs. (v) and (vi),

we get s > t

Also, a101 = a1 + 100 D

and b101 = 2100 b1

$$\therefore$$ $${a_{101}} = {a_1} + 100\left( {{{{2^{50}}{a_1} - {a_1}} \over {50}}} \right)$$

and b101 = 2100 a1

$$\Rightarrow$$ a101 = a1 + 251 a1 $$-$$ 2a1 = 251 a1 $$-$$ a1

$$\Rightarrow$$ a101 < 251 a1

and b101 > 251 a1 $$\Rightarrow$$ b101 > a101

2

IIT-JEE 2012 Paper 2 Offline

MCQ (Single Correct Answer)
Let $${a_1},{a_2},{a_3},.....$$ be in harmonic progression with $${a_1} = 5$$ and $${a_{20}} = 25.$$ The least positive integer $$n$$ for which $${a_n} < 0$$ is
A
22
B
23
C
24
D
25
3

IIT-JEE 2009

MCQ (Single Correct Answer)
In the sum of first $$n$$ terms of an A.P. is $$c{n^\angle }$$, then the sum of squares of these $$n$$ terms is
A
$${{n\left( {4{n^2} - 1} \right){c^2}} \over 6}$$
B
$${{n\left( {4{n^2} + 1} \right){c^2}} \over 3}$$
C
$${{n\left( {4{n^2} - 1} \right){c^2}} \over 3}$$
D
$${{n\left( {4{n^2} + 1} \right){c^2}} \over 6}$$
4

IIT-JEE 2008

MCQ (Single Correct Answer)
Suppose four distinct positive numbers $${a_1},\,{a_{2\,}},\,{a_3},\,{a_4}\,$$ are in G.P. Let $${b_1} = {a_1},{b_2} = {b_1} + {a_2},\,{b_3} = {b_2} + {a_{3\,\,}}\,\,\,and\,\,\,{b_4} = {b_3} + {a_4}$$.

STATEMENT-1: The numbers $${b_1},\,{b_{2\,}},\,{b_3},\,{b_4}\,$$ are neither in A.P. nor in G.P. and

STATEMENT-2 The numbers $${b_1},\,{b_{2\,}},\,{b_3},\,{b_4}\,$$ are in H.P.

A

STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is a correct explanation for
STATEMENT-1
B

STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is NOT a correct explanation for
STATEMENT-1
C
STATEMENT-1 is True, STATEMENT-2 is False
D
STATEMENT-1 is False, STATEMENT-2 is True

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