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If `a , b , c`are in A.P. `b , c , d`are in G.P. and `1/c ,1/d ,1/e`are in A.P. prove that `a , c , e`are in G.P.? |
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Answer» a, b, c are in `AP rArr 2b = a+c` ...(i) b, c, d are in GP `rArr c^(2)=bd` ...(ii) `1/c, 1/d, 1/e` are in `APrArr 2/d =(1/c+1/e)=((c+e))/(ce)` `rArr d=(2ce)/((c+e))` ...(iii) Now, `c^(2)=bd rArr c^(2) =((a+c))/2. (2ce)/((c+e))` [from (i) and (iii)] `rArr c=((a+c)e)/((c+e))` `rArr c(c+e)=(a+c)e` `rArr c^(2)=ae` `rArr` a, c, e are in GP. Hence, a, c, e are in GP. |
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