The sum of the first three terms of a G.P. is \(\frac{39}{10}\), and t

1.	The sum of the first three terms of a G.P. is \(\frac{39}{10}\), and their product is 1. Find the common ratio and the terms.
Answer» Let the three numbers be \(\frac{a}{r},a, ar\). ∴ According to the question \(\Rightarrow\) \(\frac{a}{r}\) + a+ ar = \(\frac{39}{10}\)...... (1) \(\Rightarrow\) \(\frac{a}{r}\) \(\times a \times ar = 1\) ...... (2) From 2 we get ⇒ a³ = 1 ∴ a = 1. From 1 we get \(\frac{a + ar + ar^2}{r} = \frac{39}{10}\) ⇒ 10a + 10ar + 10ar² = 39r …(3) Substituting a = 1 in 3 we get ⇒ 10(1) + 10(1)r + 10(1)r² = 39r ⇒10r² – 29r + 10 = 0 ⇒ 10r² – 25r – 4r + 10 = 0…(4) ⇒ 5r(2r – 5) – 2(2r – 5) = 0 \(\therefore\) r = \(\frac{2}{5}\) or r = \(\frac{5}{2}\) \(\therefore\) Now the equation will be ⇒ \(\cfrac{1}{\frac{2}{5}}\), 1,1\(\times\) \(\frac{2}{5}\) or \(\frac{1}{\frac{5}{2}}\),1,1 \(\times\) \(\frac{5}{2}\) ⇒ \(\frac{5}{2}, 1, \frac{2}{5}\) or \(\frac{5}{2}, 1, \frac{2}{5}\) \(\therefore\) The three numbers are \(\frac{5}{2}, 1, \frac{2}{5}\)

The sum of the first three terms of a G.P. is \(\frac{39}{10}\), and their product is 1. Find the common ratio and the terms.

Answer»

Let the three numbers be \(\frac{a}{r},a, ar\).

∴ According to the question

\(\Rightarrow\) \(\frac{a}{r}\) + a+ ar = \(\frac{39}{10}\)...... (1)

\(\Rightarrow\) \(\frac{a}{r}\) \(\times a \times ar = 1\) ...... (2)

From 2 we get

⇒ a³ = 1

∴ a = 1. From 1 we get

\(\frac{a + ar + ar^2}{r} = \frac{39}{10}\)

⇒ 10a + 10ar + 10ar² = 39r …(3)

Substituting a = 1 in 3 we get

⇒ 10(1) + 10(1)r + 10(1)r² = 39r

⇒10r² – 29r + 10 = 0

⇒ 10r² – 25r – 4r + 10 = 0…(4)

⇒ 5r(2r – 5) – 2(2r – 5) = 0

\(\therefore\) r = \(\frac{2}{5}\) or r = \(\frac{5}{2}\)

\(\therefore\) Now the equation will be

⇒ \(\cfrac{1}{\frac{2}{5}}\), 1,1\(\times\) \(\frac{2}{5}\) or \(\frac{1}{\frac{5}{2}}\),1,1 \(\times\) \(\frac{5}{2}\)

⇒ \(\frac{5}{2}, 1, \frac{2}{5}\) or \(\frac{5}{2}, 1, \frac{2}{5}\)

\(\therefore\) The three numbers are \(\frac{5}{2}, 1, \frac{2}{5}\)