Question

If 6th, 12th and 18th terms of a GP are P, Q and R respectively, then prove that P, Q and R are in G.P.

Answer 1

Let's say a,b then

GM=
ab
​
=24 or ab=576
a=
b
576
​


HM=
a+b
2ab
​
=
5
72
​


a+b
2×576
​
=
5
72
​


We get
a+b=80

On substituting the value of a,
b
576
​
+b−80=0

b
2
−80b+576=0
b=
2×1
+80±
80
2
−4×1×576
​

​


b=40±32
b=8, 72
a=72,8

Answer 2

If the common ratio of the GR is r, then

Q/P = r^6
R/Q = r^6
since there is a common ratio (R/Q = Q/P), P,Q,R form a GP

Answer 3

[AI-generated answer — Explain Bot]

To prove that P, Q, and R are in a geometric progression (GP), we need to show that the ratio of any two consecutive terms is the same.

Let's assume the common ratio of the GP is denoted by 'r'.

The 6th term of the GP, denoted by T6, can be calculated using the formula:

T6 = P * r^5 -- (1)

Similarly, the 12th term, denoted by T12, can be calculated using:

T12 = Q * r^11 -- (2)

And the 18th term, denoted by T18, can be calculated using:

T18 = R * r^17 -- (3)

To prove that P, Q, and R are in a GP, we need to show that the ratio of any two consecutive terms is the same, i.e.,

Q/P = R/Q

Let's substitute the values of T6, T12, and T18 from equations (1), (2), and (3), respectively:

(Q/P) = (Q * r^11) / (P * r^5)
= Q/P * r^6 -- (4)

(R/Q) = (R * r^17) / (Q * r^11)
= R/Q * r^6 -- (5)

From equation (4) and equation (5), we see that (Q/P) = (R/Q) = r^6.

Therefore, we have shown that the ratio of any two consecutive terms (Q/P and R/Q) is the same and equal to r^6. Hence, P, Q, and R are in a geometric progression (GP).