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Exercise 5.2 · Q7

Q.The sum of the first three terms of a G.P. is 3910\dfrac{39}{10} and their product is 1. Find the common ratio and the terms.

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Represent the three terms as a/r, a, ara/r,\,a,\,ar so the product gives aa directly, then solve a quadratic for rr.

Three consecutive G.P. terms can be written ar, a, ar\dfrac{a}{r},\,a,\,ar, with product =a3=a^3.

  1. Let terms be ar,a,ar\dfrac{a}{r},a,ar. Product =ar⋅a⋅ar=a3=1⇒a=1=\dfrac{a}{r}\cdot a\cdot ar=a^3=1\Rightarrow a=1.
  2. Sum: ar+a+ar=3910⇒1r+1+r=3910\dfrac{a}{r}+a+ar=\dfrac{39}{10}\Rightarrow\dfrac1r+1+r=\dfrac{39}{10}.
  3. So r+1r=3910−1=2910r+\dfrac1r=\dfrac{39}{10}-1=\dfrac{29}{10}.
  4. Multiply by rr: r2−2910r+1=0⇒10r2−29r+10=0r^2-\dfrac{29}{10}r+1=0\Rightarrow10r^2-29r+10=0. …

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