Let a be the first term and r be the common ratio of the G.P.
According to the given condition,
a4 = a r3 = x … (1)
a10 = a r9 = y … (2)
a16 = a r15 = z … (3)
On dividing (2) by (1), we get
\( \dfrac{y}{x}= \dfrac{ar^9}{ar^3}\)
= \( \dfrac{y}{x}=r^6\)
Now, divide equation (3) by (2).
\( \dfrac{z}{y}= \dfrac{ar^{15}}{ar^9}\)
\( \dfrac{z}{y}= r^6\)
\( \dfrac{y}{x}= \dfrac{z}{y}\)
Therefore, it is proved that x, y, z are in G. P.
Answered by Abhisek | 1 year agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.
Find the geometric means of the following pairs of numbers:
(i) 2 and 8
(ii) a3b and ab3
(iii) –8 and –2
Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).