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

a_{1} + a_{2} + a_{3} = 15

Where, the three number are: a, a + d, and a + 2d

So,

a + a + d + a + 2d = 15

3a + 3d = 15 or a + d = 5

d = 5 – a … (i)

Now, according to the question:

a + 1, a + d + 3, and a + 2d + 9

they are in GP, that is:

\(\dfrac{ (a+d+3)}{(a+1)} = \dfrac{(a+2d+9)}{(a+d+3)}\)

(a + d + 3)^{2} = (a + 2d + 9) (a + 1)

a^{2} + d^{2} + 9 + 2ad + 6d + 6a = a^{2} + a + 2da + 2d + 9a + 9

(5 – a)^{2} – 4a + 4(5 – a) = 0

25 + a^{2} – 10a – 4a + 20 – 4a = 0

a^{2} – 18a + 45 = 0

a^{2} – 15a – 3a + 45 = 0

a(a – 15) – 3(a – 15) = 0

a = 3 or a = 15

d = 5 – a

d = 5 – 3 or d = 5 – 15

d = 2 or – 10

Then,

For a = 3 and d = 2, the A.P is 3, 5, 7

For a = 15 and d = -10, the A.P is 15, 5, -5

The numbers are 3, 5, 7 or 15, 5, – 5

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