Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
125 views
in Geometric Progressions by (15.7k points)

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.

Please log in or register to answer this question.

1 Answer

0 votes
by (15.2k points)

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 \(\frac{5}{2}, 1, \frac{2}{5}\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...