If pth, qth, rth and sth terms of an A.P., be in G.P., then prove that p – q, q – r, r – s are in G.P.

Given,

pth, qth rth and sth terms of an AP are in GP .

Firstly we should find out pth, qth, rth and sth terms

Let a is the first term and d is the common difference of an AP

so, pth term = a + (p – 1)d

qth term = a + (q – 1)d

rth term = a + (r – 1)d

sth term = a + (s – 1)d

∴ [a + (p – 1)d ], [a + (q – 1)d ], [a + (r – 1)d ], [a + (s – 1)d ] are in GP

so, Let first term of GP be α and common ratio is β

Then, [a + (p – 1)d ] = α

[a + (q – 1)d ] = αβ

[a + (r – 1)d ] = αβ²

[a + (s – 1)d ] = αβ³

now, here, it is clear that α, αβ, αβ², αβ³ are in GP

NOTE: Using property of GP,we know that if a common term is multiplied with each number in a GP,series itself remains a GP

∴ α(1 – β), αβ(1 – β), αβ²(1 – β) are in GP

Where the first term is α(1 – β), and the common ratio is β

so, α(1 – β) = [a + (p – 1)d] – [a + (q – 1)d ] = (p – q)

∴ α(1 – β) = (p – q) ...... (1)

Similarly, αβ(1 – β) = αβ – αβ² = [a + (q – 1)d ] – [a + (r – 1)d] = (q – r)

∴ αβ(1 – β) = (q – r) …… (2)

And αβ²(1 – β) = αβ² – αβ³ = [α + (r – 1)d] – [α + (s – 1)d] = (r – s)

∴ αβ²(1 – β) = (r – s) …… (3)

From the above explanation, we got α(1 – β), αβ(1 – β), αβ²(1 – β) are in GP

∴ From equations (1), (2) and (3),

(p – q), (q – r), (r – s) are in GP .