Let a_1,a_2,,a_21 be an AP such that _n = 1^202mu1a_na_n + 1 = 49. If the sum of this AP is 189 , then a_6a_16 is equal to

(d): Let t_n=1a_n a_n+1=1d(a_n+1-a_n)aligned& t_n=1d(1a_n-1a_n+1) \\& _n=1^20 t_n=1d(1a_1-1a_2+1a_2-1a_3 +1a_20-1a_21) \\& =1d(1a_1-1a_21)=(a_21-a_1)d a_1 a_21=20 dd a_21 a_1=20a_1 a_21=49alignedaligned& a_1 a_21=45 \#(i) \\& 189=212(a_1+a_21) a_1+a_21=18 \\& Using (i)) \\& a_1+45a_1=18 a_1^2-18 a_1+45=0 \\& (a_1-15)(a_1-3)=0 a_1=3 or 15aligned I:

Let a_{1},a_{2},,a_{21} be an AP such that \sum_{n = 1}^{20}\mspace{2mu}\frac{1}{a_{n}a_{n + 1}} = \frac{4}{9}. If the sum of this AP is 189 , then a_{6}a_{16} is equal to: 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

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

Let a_{1},a_{2},a_{3},. be a G.P. of increasing positive numbers. Let the sum of its 6^{{th}{~}} and 8^{{th}{~}} terms be 2 and the product of its 3^{{rd}{~}} and 5^{{th}{~}} terms be 1/9. Then 6( a_{2} + a_{4} )( a_{4} + a_{6} ) is equal to

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Let \frac{1}{x_{1}},\frac{1}{x_{2}},,\frac{1}{x_{n}}( x_{i} eq 0 .\ for . \ i = 1,2,,n ) be in A.P. such that x_{1} = 4 and x_{21} = 20. If n is the least positive integer for which x_{n} > 50, then \sum_{i = 1}^{n}\mspace{2mu}( \frac{1}{x_{i}} ) is equal to

Let 3,a,b,c be in A.P. and 3,a - 1,b + 1,c + 9 be in G.P. Then, the arithmetic mean of a,b and c is

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