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_4h_7 is

(d) : Let a_2, a_3 ... a_9 are 8 airthmetic means between a_1=2 and a_10=3 Therefore a_10=10^ th term of an A.P. whose first term is a_1=2 or 3=a_1+9 d=2+ 9 d, d=(1 / 9) Now, a_4=a_1+3 d=2+3((1 / 9))=2+(1 / 3)=(7 / 3) .Again let h_2, h_3, ... ... , h_9 are 8 harmonic means between h_1=2, h_10=3. i.e., (1 / h_2), (1 / h_3), ... (1 / h_9) are 8 arithmetic means between (1 / 2) and (1 / 3) Therefore (1 / 3)=(1 / 2)+9 D \ => D=(-1 / 54) \ Therefore (1 / h_7)=(1 / h_1)+6 D=(1 / 2)-(6 / 54)\ Therefore =(1 / 2)-(1 / 9)=(7 / 18) or h_7=(18 / 7) \ Therefore a_4 h_7=((7 / 3))((18 / 7))=6

Let a_{1},a_{2}, a_{10} be in A.P. and h_{1},h_{2}, h... | 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

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