Arithmetic and geometric progressions: VJC 2024 Promo Q4

(a)

$$\begin{aligned} S_n-S_{n-1}&=75-\frac{3^{n+1}}{5^{n-2}}-\left[75-\frac{3^n}{5^{n-3}}\right]\\ u_n&=\frac{3^n}{5^{n-3}}-\frac{3^{n+1}}{5^{n-2}}\\ &=\frac{3^{n-1}}{5^{n-1}}\left(\frac3{5^{-2}}-\frac{3^2}{5^{-1}}\right)\\ &=\left(\frac35\right)^{n-1}(75-45)\\ &=30\left(\frac35\right)^{n-1} \end{aligned}$$

This is a GP with common ratio $r=\dfrac35$ and first term $a=30$

$S_\infty=\dfrac{30}{1-\frac35}=75$

OR

$S_n=75-\dfrac{3^{n+1}}{5^{n-2}}=75-27\left(\dfrac35\right)^{n-2}$

As $n\to\infty$, $\left(\dfrac35\right)^{n-2}\to0$. Hence, $S_\infty=75$

(b)

$\displaystyle\sum_{i=1}^{n}v_i=\frac{2n}{2n+1}=1-\frac1{2n+1}$

As $n\to\infty$, $\dfrac1{2n+1}\to0$. Hence, $\displaystyle\sum_{i=1}^{\infty}v_i=1$.

$\displaystyle\sum_{i=11}^{\infty}v_i=\sum_{i=1}^{\infty}v_i-\sum_{i=1}^{10}v_i=1-\frac{20}{21}=\frac1{21}$