Given:
Sum of GP = 728
Where, r = 3, a = ?
Firstly,
Tn = arn-1
486 = a3n-1
486 = \(\dfrac{ a3^n}{3}\)
486 (3) = a3n
1458 = a3n …. Equation (i)
By using the formula,
Sum of GP for n terms = \(\dfrac{ a(r^n – 1 )}{(r – 1)}\)
728 = \(\dfrac{ a (3^n – 1)}{2}\)
1456 = a3n – a … equation (2)
Subtracting equation (1) from (2) we get
1458 – 1456 = a.3n – a.3n + a
a = 2.
The first term is 2
Answered by Sakshi | 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 \( \dfrac{81}{2}\).