1

### JEE Advanced 2015 Paper 2 Offline

Numerical
The coefficient of $${x^9}$$ in the expansion of (1 + x) (1 + $${x^2)}$$ (1 + $${x^3}$$) ....$$(1 + {x^{100}})$$ is

## Explanation

Given expression is

$$E = (1 + x)(1 + {x^2})(1 + {x^3})......(1 + {x^{100}})$$

Coefficient of x9 in E

= coefficient of x9 in $$(1 + x)(1 + {x^2})(1 + {x^3})......(1 + {x^9})$$

$$\Rightarrow$$ Terms containing x9

$$= (1\,.\,{x^9} + {x^1}\,.\,{x^8} + {x^2}\,.\,{x^7} + {x^3}\,.\,{x^6} + {x^4}\,.\,{x^5} + {x^1}\,.\,{x^2}\,.\,{x^6} + {x^1}\,.\,{x^3}\,.\,{x^5} + {x^1}\,.\,{x^4}\,.\,{x^5})$$

$$\Rightarrow$$ Term containing x9 is 8x9 in E $$\Rightarrow$$ Coefficient of x9 = 8.

2

### JEE Advanced 2015 Paper 2 Offline

Numerical
Suppose that all the terms of an arithmetic progression (A.P) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is

## Explanation

$${{{S_7}} \over {{S_{11}}}} = {6 \over {11}}$$ ...... (1)

$$130 \le {t_7} \le 140$$ ........ (2)

$$\Rightarrow {{{7 \over 2}[2a + 6d]} \over {{{11} \over 2}[2a + 10d]}} = {6 \over {11}}$$

$$\Rightarrow {{a + 3d} \over {a + 5d}} = {6 \over 7}$$ ....... (3)

$$\Rightarrow {{{t_4}} \over {{4_6}}} = {6 \over 7}$$

Let $${t_4} = 6k$$, $${t_6} = 7k$$;

$$2d = k \Rightarrow d = k/2$$ and $$a + 3d = 6k$$

$$\Rightarrow a = 6k - 3k/2 = 9k/2$$

Hence, $$130 \le {t_7} \le 140$$.

$$\Rightarrow 130 \le {{9k} \over 2} + 3k \le 140$$

$$\Rightarrow 130 \le {{15k} \over 2} \le 140$$

$$\Rightarrow {{52} \over 3} \le k \in {{56} \over 3}$$

Since, $$k \in N \Rightarrow k = 18$$.

$$\Rightarrow d = {k \over 2} = {{18} \over 2} = 9$$

3

### JEE Advanced 2014 Paper 1 Offline

Numerical
Let a, b, c be positive integers such that $${b \over a}$$ is an integer. If a, b, c are in geometric progression and the arithmetic mean of a, b, c is b + 2, then the value of $${{{a^2} + a - 14} \over {a + 1}}$$ is

4

### JEE Advanced 2013 Paper 1 Offline

Numerical
A pack contains $$n$$ cards numbered from $$1$$ to $$n.$$ Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is $$1224.$$ If the smaller of the numbers on the removed cards is $$k,$$ then $$k-20=$$

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