Question

find the possible values of a ( refer to Attachment)​

Answer 1

Answer:-

$\displaystyle\sf \lim_{n\to \infty} \dfrac{\sum\limits_{r=1}^{n} r^{1/3}}{n^{7/3} \left( \sum\limits_{r=1}^{n} \dfrac{1}{(an+r)^2} \right)}$

$\displaystyle\sf \lim_{n\to \infty} \left( \dfrac{\sum\limits_{r=1}^{n} \left( \dfrac{r}{n} \right)^{1/3}}{\sum\limits_{r=1}^{n} \dfrac{1}{(a+r/n)^2}} \right)$

now, $\displaystyle\sf \dfrac{\displaystyle\sf\int\limits_0^1 x^{1/3} dx}{\displaystyle\sf\int\limits_0^1 \dfrac{1}{(a+x)^2} dx } = 54$

$\displaystyle\sf \implies \dfrac{\frac{3}{4} \bigg[ x^{4/3} \bigg]^1_0}{- \bigg[ \dfrac{1}{a+x} \bigg]^1_0 } = 54$

$\sf \implies \dfrac{\frac{3}{4}}{\frac{1}{a(a+1)}} = 54$

$\sf\implies a(a-1) = 72$

$\sf \implies a^2 + a - 72 = 0$

now use quadratic formula

$\sf x = \dfrac{-b\pm \sqrt{b^2 - 4\tt a \sf c}}{2\tt a}$

$\sf a = \dfrac{-1\pm \sqrt{1^2 -4(-72)}}{2}$

$\sf a = \dfrac{-1\pm \sqrt{289}}{2}$

$\sf a = \dfrac{-1\pm 17}{2}$

$\boxed{\sf a = -9 \ or \ 8}$