JEE Advanced Sequence and Series 1

Consider $31^x, 31^x, 31^x,\cdots (y\ times)$ and $65^y,65^y,65^y,\cdots (x\ times)$

$AM=\dfrac{(31^x+31^x+\cdots)+(65^y+65^y+\cdots)}{x+y}=\dfrac{y31^x+x65^y}{x+y}$

$GM=((31^x31^x31^x\cdots)(65^y65^y65^y\cdots))^\dfrac{1}{x+y}\\=(31^{xy}65^{xy})^\dfrac{1}{x+y}=(2015)^\dfrac{xy}{x+y}$

Now $\dfrac{1}{x}+\dfrac{1}{y}=1\Rightarrow \dfrac{x+y}{xy}=1$

$GM=2015$

By $AM-GM$ inequality,

$AM\geq GM$

$\dfrac{y31^x+x65^y}{x+y}\geq 2015\Rightarrow\dfrac{y31^x+x65^y}{xy}\geq 2015$

$\Rightarrow \dfrac{31^x}{x}+\dfrac{65^y}{y}\geq 2015$

So minimum value is 2015

question is classic As 1/x+1/y=1 so , put x=2 and y= 2,(which satisfy the given condition) Now, for two numbers 31^2 , and 65^2 Therefore by applying A.M>=G.M

31^2+65^2/2 >= sqrt (31^2 * 65^2) so, the minimum value is 31*65= 2015

The given equation is $\dfrac{1}{x} + \dfrac{1}{y} = 1$ It can be rewritten as $y = \dfrac{x}{x-1}$ Now we can write that $\alpha = \dfrac{31^{x}}{x} + 65^{\dfrac{x}{x-1}}\Bigg( 1- \dfrac{1}{x}\Bigg)$ Differentiating wrt to $x$ . $\dfrac{d\alpha}{dx} = \dfrac{31^{x} \ln31 x - 31^{x}}{x^{2}} + 65^{\dfrac{x}{x-1}} \ln65 \Bigg( \dfrac{-1}{(x-1)^{2}} \Bigg) \Bigg( \dfrac{x-1}{x} \Bigg) + \dfrac{1}{x^{2}} 65^{\dfrac{x}{x-1}} .$ When $\alpha$ is minimum, $\frac{d\alpha}{dx}$ should be $0$ . $\dfrac{31^{x} (x \ln31 -1)}{x^{2}} = 65^{\dfrac{x}{x-1}} \Bigg( \dfrac{\ln65 }{x(x-1)} -\dfrac{1}{x^{2}} \Bigg)$ Now we can replace $\frac{x}{x-1}$ by $y$ . Hence:- $31^{x} ( x\ln31 - 1) = 65^{y} (y\ln65 -1)$ These two expressions will be equal when $x\ln31 = y\ln65 \implies 31^{x} = 65^{y}$ We can again replace $y$ by $\frac{x}{x-1}$ . Hence $x\ln31 = \dfrac{x}{x-1} \ln65$ This gives $x = \dfrac{\ln (65\times 31)}{\ln 31}$ Now we know that $\alpha = \dfrac{31^{x}}{x} + \dfrac{65^{y}}{y}$ Put $31^{x} = 65^{y}$ $\alpha = 31^{x} \Bigg( \dfrac{1}{x}+\dfrac{1}{y} \Bigg)$ $\alpha = 31^{x}$ $\ln \alpha = x \ln31$ $\ln \alpha = \ln (65\times 31)$ $\alpha= 65\times 31$ $\boxed{\alpha = 2015}$

we can apply am gm in question then weighted am gm considering 31 to be x times and 65 to be y times then putting the value of x+y as xy then again applying am gm i got the answer . i am just giving only the hint . note that minimum value of xy comes out to be 4 which is helpful in solving this question

use Holder's Inequality