Let a_1,a_2, ... ,a_21 be an AP such thatn sum _n = 1^20mspace2mufrac1a_na_n + 1 = (4 / 9).n If the sum of this AP is 189 , then a_6a_16 is equal to

(d): Let t_n=frac1a_n * a_n+1=(1 / d)(a_n+1-a_n)nn n => t_n=(1 / d)((1 / a_n)-frac1a_n+1) n => sum _n=1^20 t_n=(1 / d)((1 / a_1)-(1 / a_2)+(1 / a_2)-(1 / a_3) ... +frac1a_20-frac1a_21) n =(1 / d)((1 / a_1)-frac1a_21)=frac(a_21-a_1)d a_1 a_21=frac20 dd a_21 * a_1=frac20a_1 a_21=(4 / 9)n nnn n qquad Therefore a_1 a_21=45 \ (i) n => 189=(21 / 2)(a_1+a_21) => a_1+a_21=18 n Using (i)) n => a_1+(45 / a_1)=18 => a_1^2-18 a_1+45=0 n => (a_1-15)(a_1-3)=0 => a_1=3 or 15n nnnCase I: When a_1=3nn n => a_21=15=3+20 d => d=(3 / 5) n Therefore a_6=3+5 x (3 / 5)=6 and a_16=3+15 x (3 / 5)=12 n Therefore a_6 a_16=72n nnnCase II: When a_1=15nn n => a_21=3=15+20 d => d=(-3 / 5) n Therefore a_6=15+5 x ((-3 / 5))=12 and a_16=15+15 x ((-3 / 5))=6 n Therefore a_6 a_16=72n n

Let a_{1},a_{2},,a_{21} be an AP such that \sum_{n ... | Sequence & Series (Mathematics)

(d): Let

t_n=\frac{1}{a_n \cdot a_{n+1}}=\frac{1}{d}\left(a_{n+1}-a_n\right)

\n\begin{aligned}\n& \Rightarrow t_n=\frac{1}{d}\left(\frac{1}{a_n}-\frac{1}{a_{n+1}}\right) \\\n& \Rightarrow \sum_{n=1}^{20} t_n=\frac{1}{d}\left(\frac{1}{a_1}-\frac{1}{a_2}+\frac{1}{a_2}-\frac{1}{a_3} \ldots+\frac{1}{a_{20}}-\frac{1}{a_{21}}\right) \\\n& =\frac{1}{d}\left(\frac{1}{a_1}-\frac{1}{a_{21}}\right)=\frac{\left(a_{21}-a_1\right)}{d a_1 a_{21}}=\frac{20 d}{d a_{21} \cdot a_1}=\frac{20}{a_1 a_{21}}=\frac{4}{9}\n\end{aligned}\n

\n\n\begin{aligned}\n& \qquad \therefore a_1 a_{21}=45 \#(i) \\\n& \Rightarrow 189=\frac{21}{2}\left(a_1+a_{21}\right) \Rightarrow a_1+a_{21}=18 \\\n& \text { Using }(i)) \\\n& \Rightarrow a_1+\frac{45}{a_1}=18 \Rightarrow a_1^2-18 a_1+45=0 \\\n& \Rightarrow\left(a_1-15\right)\left(a_1-3\right)=0 \Rightarrow a_1=3 \text { or } 15\n\end{aligned}\n

\n\nCase I: When $a_1=3$\n

\n\begin{aligned}\n& \Rightarrow a_{21}=15=3+20 d \Rightarrow d=\frac{3}{5} \\\n& \therefore a_6=3+5 \times \frac{3}{5}=6 \text { and } a_{16}=3+15 \times \frac{3}{5}=12 \\\n& \therefore a_6 a_{16}=72\n\end{aligned}\n

\n\nCase II: When $a_1=15$\n

\n\begin{aligned}\n& \Rightarrow a_{21}=3=15+20 d \Rightarrow d=\frac{-3}{5} \\\n& \therefore a_6=15+5 \times\left(\frac{-3}{5}\right)=12 \text { and } a_{16}=15+15 \times\left(\frac{-3}{5}\right)=6 \\\n& \therefore a_6 a_{16}=72\n\end{aligned}\n

Explanation

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