Question

if pth, qth, rthand sth terms of an A.P. are in G.P.,show that (p-q),(q-r) and (r-s) are also in G.P.

Answer 1

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 is 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
∴ α(1 - β) , αβ(1 - β), αβ²(1 - β) are in GP
Where first term is α(1 - β) and common ratio is β

so, α(1 - β) = [a + (p -1)d] - [a + (q - 1)d ] = (p - q) ----(1)
αβ(1 - β) = αβ - αβ² = [a + (q - 1)d ] - [a + (r - 1)d] = (q - r) ------(2)
αβ²(1 - β) = αβ² - αβ³ = [α + (r - 1)d] - [α + (s - 1)d] = (r - s) ------(3)

From 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
Hence proved//

Answer 2

Answer:

Here,

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 is 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

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

Where first term is α(1 - β) and common ratio is β

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

αβ(1 - β) = αβ - αβ² = [a + (q - 1)d ] - [a + (r - 1)d] = (q - r) ------(2)

αβ²(1 - β) = αβ² - αβ³ = [α + (r - 1)d] - [α + (s - 1)d] = (r - s) ------(3)

From 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

Hope this helps you.