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)
a2 + d2 + 16 + 8a + 2ad + 8d = a2 + 4a + 2da + 36a + 144 + 8d
d2 – 32a – 128
(6 – a)2 – 32a – 128 = 0
36 + a2 – 12a – 32a – 128 = 0
a2 – 44a – 92 = 0
a2 – 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) a3b and ab3
(iii) –8 and –2
Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).