Let’s consider the terms in A.P. to be a, a + d, a + 2d, a + 3d, … a + (n – 2) d, a + (n – 1)d.

From the question, we have

Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d

Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]

= 4a + (4n – 10) d

Then according to the given condition,

4a + 6d = 56

4(11) + 6d = 56 [Since a = 11 (given)]

6d = 12

d = 2

Hence, 4a + (4n –10) d = 112

4(11) + (4n – 10)2 = 112

(4n – 10)2 = 68

4n – 10 = 34

4n = 44

n = 11

Therefore, the number of terms of the A.P. is 11.

Answered by Pragya Singh | 2 years 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 \( \dfrac{81}{2}\).