Question

The numbers 2/1/3,h,k,7/7/8 firm a geometric progression. find the value of h and of k

Answer 1

$\text{We have }\ 2\dfrac{1}{3},\ h,\ k,\ 7\dfrac{7}{8}.\\\\\text{Convert the mixed numbers to the improper fractions}\\\\2\dfrac{1}{3}=\dfrac{2\cdot3+1}{3}=\dfrac{7}{3}\\\\7\dfrac{7}{8}=\dfrac{7\cdot8+7}{8}=\dfrac{63}{8}\\\\\text{If}\ a_1,\ a_2,\ a_3,\ a_4\ \text{is a geometric seqence, then:}\\\\\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=r\\\\a_2=a_1r\\\\\text{and}\ \dfrac{a_4}{a_1}=r^3\\\\\text{Substitute:}$

$r^3=\dfrac{\frac{63}{8}}{\frac{7}{3}}\\\\r^3=\dfrac{63}{8}\cdot\dfrac{3}{7}\\\\r^3=\dfrac{9\cdot3}{8\cdot1}\\\\r^3=\dfrac{27}{8}\to r=\sqrt[3]{\dfrac{27}{8}}\\\\r=\dfrac{3}{2}\\\\h=2\dfrac{1}{3}\cdot\dfrac{3}{2}=\dfrac{7}{3}\cdot\dfrac{3}{2}=\dfrac{7}{2}\\\\k=\dfrac{7}{2}\cdot\dfrac{3}{2}=\dfrac{21}{4}\\\\Answer:\ \boxed{h=\dfrac{7}{2}=3\dfrac{1}{2}\ and\ k=\dfrac{21}{4}=5\dfrac{1}{4}}$