There is a set of four numbers p q r and s respectively in such a manner that first three are in G P...........

Quantitative Aptitude ( CCS) Arithmetic progression

There is a set of four numbers p, q, r and s respectively in such a manner that first three are in G.P. and the last three are in A.P. with a difference of 6. If the first and the fourth numbers are the same, find the value of p.

–4

–24

छोटे Courses बड़े Results

Master Arithmetic progression

Prepare through Micro-courses

Try Now

Explanation:

Let the first three numbers be a /r, a, ar.
This also implies the 4th number will be (ar + 6).
Out of these the first and the fourth numbers are same i.e. a/r = ar + 6 (ii).
It can also be said that ar = a + 6
⇒ ar – a = 6 ⇒ a(r – 1) = 6 or a = 6/(r – 1) (iii).
Put the value of a from (iii) in (ii), we get the quadratic equation as 2r²– r – 1 = 0.
Solving it, we get the value of r as – 1/2 or 1/2.
Putting r as – 1/2, we get the value of a is – 4 and putting r as 1/2 we get the value of a as – 12.
The first term of the GP becomes 8 and – 24 respectively.
Similarly, the third term of the GP becomes 2 and – 6 respectively.
The two possible GP series are 8, -4, 2 and –24, - 12, - 6.
The last terms of these 2 series become 8 and 0 respectively.
See in the second possible series, the first and the fourth numbers are not same, as specified in the question. Thus the only possible for numbers are 8, - 4, 2, 8,