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Worked Examples · Example 21

Q.The sum of three consecutive terms of a G.P. is 26 and their product is 216. Find the common ratio and the terms.

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✓ Free question

Write the three terms as ar,a,ar\tfrac{a}{r},a,ar; the product gives aa directly, then the sum gives a quadratic in rr.

[!FORMULA] Three consecutive terms of a G.P. with common ratio rr can be written symmetrically as ar, a, ar\dfrac{a}{r},\,a,\,ar, so their product is a3a^3 (the middle term cubed) and their sum is a(1r+1+r)a\left(\dfrac1r+1+r\right).

  1. Let the terms be ar,a,ar\dfrac{a}{r}, a, ar.
  2. Product: ar⋅a⋅ar=a3=216 ⇒ a=6\dfrac{a}{r}\cdot a\cdot ar=a^3=216\ \Rightarrow\ a=6.
  3. Sum: ar+a+ar=26 ⇒ 6r+6+6r=26\dfrac{a}{r}+a+ar=26\ \Rightarrow\ \dfrac{6}{r}+6+6r=26.
  4. Simplify: 6r+6r=20 ⇒ 3r+3r=10\dfrac6r+6r=20\ \Rightarrow\ \dfrac{3}{r}+3r=10 (dividing by 2).
  5. Multiply through by rr: 3+3r2=10r ⇒ 3r2−10r+3=03+3r^2=10r\ \Rightarrow\ 3r^2-10r+3=0.
  6. Solve using the quadratic formula: r=10±100−366=10±86r=\dfrac{10\pm\sqrt{100-36}}{6}=\dfrac{10\pm8}{6}, so r=3r=3 or r=13r=\dfrac13.
  7. For r=3r=3: terms are 63,6,6(3)=2,6,18\dfrac{6}{3},6,6(3)=2,6,18.
  8. For r=13r=\tfrac13: terms are 18,6,218,6,2 (the same three numbers in reverse order).
  9. Check: 2+6+18=262+6+18=26 ✓ and 2×6×18=2162\times6\times18=216 ✓.
✓Final answer

Common ratio r=3r=3 (or 13\tfrac13); the terms are 2,6,182, 6, 18

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