Given that the sum of some terms in a G.P is 315.
Let the number of terms be n.
We know that, sum of terms is
\( s_n=\dfrac{a(1-r^n)}{r-1}\)
Given that the first term a is 5 and common ratio r is 2.
\(315=\dfrac{5(2^n-1)}{2-1}\)
2n-1=63
2n = 64 = (2)6
n=6
Hence, the last term of the G.P = 6th term = ar6 – 1 = (5)(2)5 = (5)(32) = 160
Therefore, the last term of the G.P. is 160.
Answered by Pragya Singh | 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 \( \dfrac{81}{2}\).