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
0 votes
207 views
in Mathematics by (85.8k points)
closed by
A geometric progression (GP) consists of 200 terms. If the sum of odd terms of the GP is m, and the sum of even terms of the GP is n, then what is its common ratio?
1. m/n
2. n/m
3. m + (n/m)
4. n + (m/n)

1 Answer

0 votes
by (96.5k points)
selected by
 
Best answer
Correct Answer - Option 2 : n/m

Concept:

If a1, a2, ..., an are in GP then common ratio is given by: r = ai + 1 / ai ∀ i =1, 2, …., n - 1.

The nth term in a GP is given by: an = arn - 1, where a is the first term and r is the common ratio.

Calculation:

Given: The sum of odd terms of the GP is m and sum of even terms of the GP is n.

\(\Rightarrow \frac{{Sum\;of\;even\;terms\;of\;GP}}{{Sum\;of\;odd\;terms\;of\;GP}} = \frac{{\left( {ar + a{r^3} + \ldots + a{r^{199}}} \right)}}{{\left( {a + a{r^2} + \ldots + a{r^{198}}} \right)}} = \frac{n}{m}\)

\(\Rightarrow \frac{{Sum\;of\;even\;terms\;of\;GP}}{{Sum\;of\;odd\;terms\;of\;GP}} = \frac{{ar \times \left( {1 + {r^2} + \ldots + {r^{198}}} \right)}}{{a \times \left( {1 + {r^2} + \ldots + {r^{198}}} \right)}} = \frac{n}{m}\)

⇒ r = n / m.

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.

Categories

...