Mathematics - Sequence & Series Question with Solution | TestHub

(A) $b-a=c-b$ and $(c-b{)}^2=a(b-a)$ give $b=2a,c=3a$; then $\frac{a^3+b^3+c^3}{3abc}=\frac{1+8+27}{18}=2$. (B) With $y=aR$, $z=a{R}^2,b=a{R}^3,x=\frac{a+b}{2}:\frac{y^3+z^3}{xyz}=\frac{a^3{R}^3\left(1+R^3\right)}{aR\cdot \frac{a\left(1+R^3\right)}{2}\cdot a{R}^2}=2$. (C) Pairing: $a_1^2-a_2^2+\cdots -a_{50}^2=-d\left(a_1+a_2+\cdots +a_{50}\right)=-25d\left(a_1+a_{50}\right)$; with $d=\frac{a_{50}-a_1}{49}$ this equals $\frac{25}{49}\left(a_1^2-a_{50}^2\right)$; direct computation gives $\left(\frac{25}{49}\right)\left(a_1^2-a_{50}^2\right)={\left(\frac{5}{7}\right)}^2\left(a_1^2-a_{50}^2\right)$; comparing with ${\left(\frac{5}{7}\right)}^n\left(a_1^2-a_{50}^2\right)$ gives $n=2$. (D) ${\tan }^{-1}\frac{1}{2r^2}={\tan }^{-1}(2r+1)-{\tan }^{-1}(2r-1)$; the sum telescopes to ${\tan }^{-1}(2n+1)-{\tan }^{-1}1\to \frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4};\tan \frac{\pi }{4}=1$.