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Question Number 56421 by gunawan last updated on 16/Mar/19

$$\mathrm{Let}{a}_1,a_2,...,a_{10}\mathrm{be}\mathrm{in}\mathrm{AP}\mathrm{and}{h}_1,h_2,...,h_{10}$$$$\mathrm{be}\mathrm{in}\mathrm{HP}.\mathrm{If}{a}_1=h_1=2\mathrm{and}{a}_{10}=h_{10}=3,$$$$\mathrm{then}{a}_4{h}_7\mathrm{is}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19

$$a_1=2\to a_{10}=a_1+9d\to$$$$3=2+9d\to d=\frac{1}{9}$$$$a_4=a_1+3d\to 2+\frac{3}{9}\to \frac{7}{3}$$$$\frac{1}{h_{10}}=\frac{1}{h_1}+9d_1\to \frac{1}{3}=\frac{1}{2}+9d_1$$$$\frac{1}{3}-\frac{1}{2}=9d_1$$$$d_1=\frac{-1}{54}$$$$\frac{1}{h_7}=\frac{1}{h_1}+6\times d_1\to \frac{1}{h_7}=\frac{1}{2}-\frac{6}{54}\to \frac{27-6}{54}=\frac{21}{54}$$$$a_4{h}_7\to \frac{7}{3}\times \frac{54}{21}\to 6$$$$$$

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