First term, a = 25 and

Common difference, d = 22 – 25 = -3

Also given, sum of certain number of terms of the A.P. is 116

The number of terms be n

So, we have

S_{n} = \( \dfrac{n}{2}\) [2a + (n-1)d] = 116

116 = \( \dfrac{n}{2}\) [2(25) + (n-1)(-3)]

116 x 2 = n [50 – 3n + 3]

232 = n [53 – 3n]

232 = 53n – 3n^{2}

3n^{2} – 53n + 232 = 0

3n^{2} – 24n – 29n+ 232 = 0

3n(n – 8) – 29(n – 8) = 0

(3n – 29) (n – 8) = 0

Hence,

n =\( \dfrac{29}{3}\) or n = 8

As n can only be an integral value, n = 8

Thus, 8^{th} term is the last term of the A.P.

a_{8} = 25 + (8 – 1)(-3)

= 25 – 21

= 4

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 \( \dfrac{81}{2}\).