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1

### JEE Advanced 2020 Paper 1 Offline

Numerical
Let a1, a2, a3, .... be a sequence of positive integers in arithmetic progression with common difference 2. Also, let b1, b2, b3, .... be a sequence of positive integers in geometric progression with common ratio 2. If a1 = b1 = c, then the number of all possible values of c, for which the equality 2(a1 + a2 + ... + an) = b1 + b2 + ... + bn holds for some positive integer n, is ...........

## Explanation

Given arithmetic progression of positive integers terms a1, a2, a3, ..... having common difference '2' and geometric progression of positive integers
terms b1, b2, b3, .... having common ratio '2' with a1 = b1 = c, such that
2(a1 + a2 + a3 + ... + an) = b1 + b2 + b3 + ... + bn

$$\Rightarrow 2 \times {n \over 2}[2C + (n - 1)2] = C\left( {{{{2^n} - 1} \over {2 - 1}}} \right)$$

$$\Rightarrow 2nC + 2{n^2} - 2n = {2^n}.C - C$$

$$\Rightarrow C[{2^n} - 2n - 1] = 2{n^2} - 2n$$

$$\because$$ $$C \in N \Rightarrow 2{n^2} - 2n \ge {2^n} - 2n - 1$$

$$\Rightarrow 2{n^2} + 1 \ge {2^n} \Rightarrow n \le 6$$

and, also C > 0 $$\Rightarrow$$ n > 2

$$\therefore$$ The possible values of n are 3, 4, 5, 6

So, at $$n = 3,\,C = {{(2 \times 9) - 6} \over {8 - 6 - 1}} = 12$$

at, $$n = 4,\,C = {{32 - 8} \over {16 - 8 - 1}} = {{24} \over 9} = {8 \over 3} \notin N$$

at, $$n = 5,\,C = {{50 - 10} \over {32 - 10 - 1}} = {{40} \over {21}} \notin N$$

and at, $$n = 6,\,C = {{72 - 12} \over {64 - 12 - 1}} = {{60} \over {51}} \notin N$$

$$\therefore$$ The required value of C = 12 for n = 3
2

### JEE Advanced 2020 Paper 1 Offline

Numerical
Let m be the minimum possible value of $${\log _3}({3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}})$$, where $${y_1},{y_2},{y_3}$$ are real numbers for which $${{y_1} + {y_2} + {y_3}}$$ = 9. Let M be the maximum possible value of $$({\log _3}{x_1} + {\log _3}{x_2} + {\log _3}{x_3})$$, where $${x_1},{x_2},{x_3}$$ are positive real numbers for which $${{x_1} + {x_2} + {x_3}}$$ = 9. Then the value of $${\log _2}({m^3}) + {\log _3}({M^2})$$ is ...........

## Explanation

For real numbers y1, y2, y3, the quantities $${{3^{{y_1}}}}$$, $${{3^{{y_2}}}}$$ and $${{3^{{y_3}}}}$$ are positive real numbers, so according to the AM-GM inequality, we have

$${{{3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}}} \over 3} \ge {({3^{{y_1}}}\,.\,{3^{{y_2}}}\,.\,{3^{{y_3}}})^{{1 \over 3}}}$$

$$\Rightarrow {3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}} \ge 3{({3^{{y_1}}}\,.\,{3^{{y_2}}}\,.\,{3^{{y_3}}})^{{1 \over 3}}}$$

On applying logarithm with base '3', we get

$${\log _3}({3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}}) \ge \left[ {1 + {1 \over 3}({y_1} + {y_2} + {y_3}} \right.)]$$

= 1 + 3

= 4

{$$\because$$ $${{y_1} + {y_2} + {y_3}}$$ = 9}

$$\therefore$$ m = 4

Now, for positive real numbers x1, x2 and x3 according to AM-GM inequality, we have

$${{{x_1} + {x_2} + {x_3}} \over 3} \ge {({x_1}{x_2}{x_3})^{{1 \over 3}}}$$

On applying logarithm with base '3', we get

$${\log _3}\left( {{{{x_1} + {x_2} + {x_3}} \over 3}} \right) \ge {1 \over 3}$$$$({\log _3}{x_1} + {\log _3}{x_2} + {\log _3}{x_3})$$

$$\Rightarrow$$ $$1 \ge {1 \over 3}\left( {{{\log }_3}{x_1} + {{\log }_3}{x_2} + {{\log }_3}{x_3}} \right)$$

{$$\because$$ x1 + x2 + x3 = 9}

$$\therefore$$ M = 3

Now, $${\log _2}({m^3}) + {\log _3}({M^2})$$

$$= 3lo{g_2}(4) + 2lo{g_2}(3)$$ = 6 + 2 = 8
3

### JEE Advanced 2019 Paper 1 Offline

Numerical
Let AP(a; d) denote the set of all the terms of an infinite arithmetic progression with first term a and common difference d > 0. If $$AP(1;3) \cap AP(2;5) \cap AP(3;7)$$ = AP(a ; d), then a + d equals ..............

## Explanation

Given that, AP(a ; d) denote the set of all the terms of an infinite arithmetic progression with first term 'a' and common difference d > 0.

Now, let mth term of first progression

$$AP(1;3) = 1 + (m - 1)3 = 3m - 2$$ .... (i)

and nth term of progression

$$AP(2;5) = 2 + (n - 1)5 = 5m - 3$$ .... (ii)

and rth term of third progression

$$AP(3;7) = 3 + (r - 1)7 = 7m - 4$$ .... (iii) are equal.

Then, $$3m - 2 = 5n - 3 = 7r - 4$$

Now, for $$AP(1;3) \cap AP(2;5) \cap AP(3;7)$$,

the common terms of first and second progressions, $$m = {{5n - 1} \over 3}$$

$$\Rightarrow$$ n = 2, 5, 11, ...

and the common terms of second and the third progressions,

$$r = {{5n + 1} \over 7}$$ $$\Rightarrow$$ n = 4, 11, ....

Now, the first common term of first, second and third progressions (when n = 11), so

a = 2 + (11 - 1)5 = 52

and d = LCM (3, 5, 7) = 105

So, $$AP(1;3) \cap AP(2;5) \cap AP(3;7)$$ = AP(52; 105)

So, a = 52 and d = 105

$$\Rightarrow$$ a + d = 157.00
4

### JEE Advanced 2018 Paper 1 Offline

Numerical
Let X be the set consisting of the first 2018 terms of the arithmetic progression 1, 6, 11, ...., and Y be the set consisting of the first 2018 terms of the arithmetic progression 9, 16, 23, .... . Then, the number of elements in the set X $$\cup$$ Y is .........

## Explanation

Here, X = {1, 6, 11, ....., 10086}

[$$\because$$ an = a + (n $$-$$ 1)d]

and Y = {9, 16, 23, ..., 14128}

X $$\cap$$ Y = {16, 51, 86, ...}

tn of X $$\cap$$ Y is less than or equal to 10086

$$\therefore$$ tn = 16 + (n $$-$$ 1) 35 $$\le$$ 10086

$$\Rightarrow$$ n $$\le$$ 288.7

$$\therefore$$ n = 288

$$\because$$ n(X $$\cap$$ Y) = n(X) + n(Y) $$-$$ n(X $$\cap$$ Y)

$$\therefore$$ n(X $$\cap$$ Y) = 2018 + 2018 $$-$$ 288 = 3748

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