Given, the k^{th} term of the A.P. is 5k + 1.

k^{th} term = a_{k} = 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) **a^{3}b and ab^{3}

**(iii) **–8 and –2

Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).