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

Let a_1 , a_2  , ..., a_(10)  be in AP and h_1 , h_2 ,..., h_(10)   be in HP. If  a_1 = h_1 =2  and  a_(10) = h_(10) =3,  then a_4 h_7   is

$$\mathrm{Let}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} \:,\:...,\:{a}_{\mathrm{10}} \:\mathrm{be}\:\mathrm{in}\:\mathrm{AP}\:\mathrm{and}\:{h}_{\mathrm{1}} ,\:{h}_{\mathrm{2}} ,...,\:{h}_{\mathrm{10}} \\ $$$$\mathrm{be}\:\mathrm{in}\:\mathrm{HP}.\:\mathrm{If}\:\:{a}_{\mathrm{1}} =\:{h}_{\mathrm{1}} =\mathrm{2}\:\:\mathrm{and}\:\:{a}_{\mathrm{10}} =\:{h}_{\mathrm{10}} =\mathrm{3}, \\ $$$$\mathrm{then}\:{a}_{\mathrm{4}} {h}_{\mathrm{7}} \:\:\mathrm{is} \\ $$

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

a_1 =2   →a_(10) =a_1 +9d→  3=2+9d→d=(1/9)  a_4 =a_1 +3d→2+(3/9)→(7/3)  (1/h_(10) )=(1/h_1 )+9d_1 →(1/3)=(1/2)+9d_1   (1/3)−(1/2)=9d_1   d_1 =((−1)/(54))  (1/h_7 )=(1/h_1 )+6×d_1 →(1/h_7 )=(1/2)−(6/(54))→((27−6)/(54))=((21)/(54))  a_4 h_7 →(7/3)×((54)/(21))→6

$${a}_{\mathrm{1}} =\mathrm{2}\:\:\:\rightarrow{a}_{\mathrm{10}} ={a}_{\mathrm{1}} +\mathrm{9}{d}\rightarrow \\ $$$$\mathrm{3}=\mathrm{2}+\mathrm{9}{d}\rightarrow{d}=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${a}_{\mathrm{4}} ={a}_{\mathrm{1}} +\mathrm{3}{d}\rightarrow\mathrm{2}+\frac{\mathrm{3}}{\mathrm{9}}\rightarrow\frac{\mathrm{7}}{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{{h}_{\mathrm{10}} }=\frac{\mathrm{1}}{{h}_{\mathrm{1}} }+\mathrm{9}{d}_{\mathrm{1}} \rightarrow\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{9}{d}_{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{9}{d}_{\mathrm{1}} \\ $$$${d}_{\mathrm{1}} =\frac{−\mathrm{1}}{\mathrm{54}} \\ $$$$\frac{\mathrm{1}}{{h}_{\mathrm{7}} }=\frac{\mathrm{1}}{{h}_{\mathrm{1}} }+\mathrm{6}×{d}_{\mathrm{1}} \rightarrow\frac{\mathrm{1}}{{h}_{\mathrm{7}} }=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{6}}{\mathrm{54}}\rightarrow\frac{\mathrm{27}−\mathrm{6}}{\mathrm{54}}=\frac{\mathrm{21}}{\mathrm{54}} \\ $$$${a}_{\mathrm{4}} {h}_{\mathrm{7}} \rightarrow\frac{\mathrm{7}}{\mathrm{3}}×\frac{\mathrm{54}}{\mathrm{21}}\rightarrow\mathrm{6} \\ $$$$ \\ $$

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