Given, the kth term of the A.P. is 5k + 1.
kth term = ak = a + (k – 1)d
And,
a + (k – 1)d = 5k + 1
a + kd – d = 5k + 1
On comparing the coefficient of k, we get d = 5
a – d = 1
a – 5 = 1
⇒ a = 6
= \( \dfrac{n}{2}[2a+(n-1)d]\)
= \( \dfrac{n}{2}[2(6)+(n-1)(5)]\)
= \( \dfrac{n}{2}[12+5n-5]\)
= \( \dfrac{n}{2}(5n+7)\)
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 \( \dfrac{81}{2}\).