Question

class 11th
please help me​

Answer 1

Step-by-step explanation:

$\begin{array}{l} \color{red} \tt(a) \text{We have, } a_{n}=4 n-3 \cdot a_{17}=4(17)-3=68-3=65 \implies a_{24}=4(24)-3=93 . \\ \\ \color{blue} \tt (b) \text{ We have,} \tt a_{n}=\dfrac{n^{2}}{2^{n} \cdot a_{7}}=\dfrac{7^{2}}{2^{7}} \qquad \qquad \Rightarrow a_{7}=\dfrac{49}{128} \\ \\ \tt \color{darkgreen}(c) a_{n}=(-1)^{n-1} \cdot n^{3} \cdot a_{9}=(-1)^{8} \cdot 9^{3} \quad \Rightarrow \quad a_{9}=729 \\ \\ \tt \color{darkviolet}(d) a_{n}=\dfrac{n(n-2)}{n+3} \cdot a_{20}=\dfrac{20(20-2)}{20+3} \qquad \Rightarrow a_{10}=\dfrac{360}{23} \end{array}$