Let a_{1},a_{2}, a_{10} be in A.P. and h_{1},h_{2}, h_{10} be in H.P. If a_{1} = h_{1} = 2 and a_{10} = h_{10} = 3, then a_{4}h_{7} is: Sequence & Series (Mathematics)

(d) : Let $a_2, a_3 \ldots a_9$ are 8 airthmetic means between $\\a_1=2$ and $a_{10}=3\\$ $\therefore a_{10}=10^{\text {th }}$ term of an A.P. whose first term is $a_1=2$ or $\\3=a_1+9 d=2+ 9 d, d=\frac{1}{9}\\$ Now, $a_4=a_1+3 d=2+3\left(\frac{1}{9}\right)=2+\frac{1}{3}=\frac{7}{3}\\$ .Again let $h_2, h_3, \ldots \ldots, h_9$ are 8 harmonic means between $\\h_1=2, h_{10}=3$ . i.e., $\frac{1}{h_2}, \frac{1}{h_3}, \ldots \frac{1}{h_9}\\$ are 8 arithmetic means between $\frac{1}{2}$ and $\frac{1}{3}\\$ $\therefore \frac{1}{3}=\frac{1}{2}+9 D \\ \Rightarrow D=\frac{-1}{54} \\ \therefore \frac{1}{h_7}=\frac{1}{h_1}+6 D=\frac{1}{2}-\frac{6}{54}\\ \therefore=\frac{1}{2}-\frac{1}{9}=\frac{7}{18} \text { or } h_7=\frac{18}{7} \\ \therefore a_4 h_7=\left(\frac{7}{3}\right)\left(\frac{18}{7}\right)=6$

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

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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

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