Let ${\left(2x^2+3x+4\right)}^{10}={\sum }_{r=0}^{20}?a_r{x}^r$ . Then $\frac{a_7}{a_{13}}$ is equal to

$1+\left(1-2^2.1\right)+\left(1-4^2.3\right)+\dots \dots .+\left(1-{20}^2.19\right)$

$=\alpha -220\beta$

$=11-\left(2^2\cdot 1+4^2\cdot 3+\dots ..+{20}^2\cdot 19\right)$

$=11-2^2\cdot {\sum }_{r=1}^{10}?r^2(2r-1)=11-4\left(\frac{{110}^2}{2}-35\times 11\right)$

$=11-220\left(103\right)$

$\Rightarrow \alpha =11,\beta =103$

$x\frac{dy}{dx}-y=x^2(x\mathrm{c}\mathrm{o}\mathrm{s}?x+\mathrm{s}\mathrm{i}\mathrm{n}?x),x>0$

$\frac{dy}{dx}-\frac{y}{x}=x(x\mathrm{c}\mathrm{o}\mathrm{s}?x+\mathrm{s}\mathrm{i}\mathrm{n}?x)\Rightarrow \frac{dy}{dx}+Py=Q$

So, I.F. $=e^{\int -\frac{1}{x}dx}\frac{1}{\left|x\right|}=\frac{1}{x}(x>0)$

Thus, $\frac{y}{x}=\int \frac{1}{x}\left(x\right(x\mathrm{c}\mathrm{o}\mathrm{s}?x+\mathrm{s}\mathrm{i}\mathrm{n}?x\left)\right)dx$

$\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{x}+\mathrm{C}$

$?\mathrm{y}\left(\pi \right)=\pi \Rightarrow \mathrm{C}=1$

So, $y=x^2\mathrm{s}\mathrm{i}\mathrm{n}?x+x\Rightarrow (y{)}_{\pi /2}=\frac{{\pi }^2}{4}+\frac{\pi }{2}$

Also, $\frac{dy}{dx}=x^2\mathrm{c}\mathrm{o}\mathrm{s}?x+2x\mathrm{s}\mathrm{i}\mathrm{n}?x+1$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-x^2\mathrm{s}\mathrm{i}\mathrm{n}?x+4x\mathrm{c}\mathrm{o}\mathrm{s}?x+2\mathrm{s}\mathrm{i}\mathrm{n}?x$

$a_1,a_2,a_3......$ are in A.P. (Let common difference is d₁)

$b_1,b_2,b_3....$  are in A.P. (Let common difference is d₂)

and a₁2, a₁₀ = 3, a₁b₁ = 1 = a₁₀b₁₀

$?a_1{b}_1=1$               

$\therefore b_1=\frac{1}{2}$

$a_{10}{b}_{10}=1$               

$\therefore b_{10}=\frac{1}{3}$              2026 Info edge India Ltd. All rights reserved.