Final answer:
To prove the differences (p-q), (q-r), and (r-s) form a geometric progression, we start by defining the terms of the AP and use the condition that they form a GP. Upon expressing each term and simplifying, we show that their consecutive differences are constant, and thus their ratio is same, proving that these differences are indeed in GP.
Step-by-step explanation:
To show that if the pth, qth, rth, and sth terms of an arithmetic progression (AP) are in geometric progression (GP), then the differences (p-q), (q-r), and (r-s) are also in GP, we will use the properties of AP and GP.
Let the AP be represented by a, a+d, a+2d, ..., where 'a' is the first term and 'd' is the common difference. The pth, qth, rth, and sth terms can be expressed as:
- a + (p-1)d
- a + (q-1)d
- a + (r-1)d
- a + (s-1)d
For these terms to be in GP, each term should be multiplied by a common ratio, 'r', to get the next term. Therefore, we can write:
(a + (q-1)d) / (a + (p-1)d) = (a + (r-1)d) / (a + (q-1)d) = (a + (s-1)d) / (a + (r-1)d)
If we simplify the above equations, we can eliminate 'a' and 'd', and then we get:
(q-p) = (r-q) = (s-r)
These differences form an AP, which means their differences are also constant. If we take successive terms of these differences and divide them, we will have:
(q-p) / (r-q) = (r-q) / (s-r)
Since they give a constant ratio, we conclude that (p-q), (q-r), and (r-s) are in GP.