Let the first term of an A.P. be ‘a’ and its common difference be‘d’.

b = a + d; c = a + 2d.

Given:

a + b + c = 18

3a + 3d = 18 or a + d = 6.

d = 6 – a … (i)

Now, according to the question:

a + 4, a + d + 4, and a + 2d + 36

they are now in GP, that is:

\(\dfrac{ (a+d+4)}{(a+4)}\) = \( \dfrac{(a+2d+36)}{(a+d+4)}\)

(a + d + 4)^{2} = (a + 2d + 36)(a + 4)

a^{2} + d^{2} + 16 + 8a + 2ad + 8d = a^{2} + 4a + 2da + 36a + 144 + 8d

d^{2} – 32a – 128

(6 – a)^{2} – 32a – 128 = 0

36 + a^{2} – 12a – 32a – 128 = 0

a^{2} – 44a – 92 = 0

a^{2} – 46a + 2a – 92 = 0

a(a – 46) + 2(a – 46) = 0

a = – 2 or a = 46

d = 6 –a

d = 6 – (– 2) or d = 6 – 46

d = 8 or – 40

Then,

For a = -2 and d = 8, the A.P is -2, 6, 14

For a = 46 and d = -40, the A.P is 46, 6, -34

The numbers are – 2, 6, 14 or 46, 6, – 34

Answered by Aaryan | 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 \( \dfrac{81}{2}\).