Aspire's Library
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
The sum of the first three terms of a G.P. is S and their product is 27. Then all such S lie in :
Solution
Let the G.P. be \(a,\ ar,\ ar^2\).
Given:
$$S = a(1 + r + r^2)$$ and $$a^3 r^3 = 27 \Rightarrow (ar)^3 = 27 \Rightarrow ar = 3.$$
Hence, $$a = \frac{3}{r}.$$
Substitute in \(S\): $$ S = \frac{3}{r}(1 + r + r^2) = 3\left(r + \frac{1}{r} + 1\right) $$
For real \(r \ne 0\):
If \(r > 0,\) then $(r + \frac{1}{r} \ge 2 \Rightarrow S \ge 3(2 + 1) = 9)$
If $(r < 0)$ then $(r + \frac{1}{r} \le -2 \Rightarrow S \le 3(-2 + 1) = -3)$
Given:
$$S = a(1 + r + r^2)$$ and $$a^3 r^3 = 27 \Rightarrow (ar)^3 = 27 \Rightarrow ar = 3.$$
Hence, $$a = \frac{3}{r}.$$
Substitute in \(S\): $$ S = \frac{3}{r}(1 + r + r^2) = 3\left(r + \frac{1}{r} + 1\right) $$
For real \(r \ne 0\):
If \(r > 0,\) then $(r + \frac{1}{r} \ge 2 \Rightarrow S \ge 3(2 + 1) = 9)$
If $(r < 0)$ then $(r + \frac{1}{r} \le -2 \Rightarrow S \le 3(-2 + 1) = -3)$
Hence all possible values of \(S\) lie in the intervals:
$$ \boxed{S \in (-\infty,\ -3]\ \cup\ [9,\ \infty)} $$
$$ \boxed{S \in (-\infty,\ -3]\ \cup\ [9,\ \infty)} $$
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
