Question

The pth,2pth and 4pth terms of an ap are in gp then the common ratio of gp is

Answer 1

common ratio of GP is 2 if The pth,2pth and 4pth terms of an AP are in GP

Step-by-step explanation:

pth Term of an AP = a + (p - 1)d

2pth term of AP = a + (2p - 1)d

4pth term of AP = a + (4p - 1)d

pth, 2pth and 4pth terms of an ap are in GP

=>  (a + (p - 1)d)(a + (4p - 1)d)  = (a + (2p - 1)d)²

=> a²  + ad(p-1 + 4p - 1) + d²(p-1)(4p-1)  = a² + d²(2p-1)² + 2ad(2p-1)

=>  ad(5p - 2)  + d²(4p² -5p + 1) = d²(4p² + 1 - 4p) + ad(4p - 2)

=> adp = d²(p)

=> a = d

pth term  = d + ( p - 1)d =  pd

2pth term = 2pd

4pth Term = 4pd

Common Ratio =  2pd/pd = 4pd/2pd  = 2

common ratio of gp is 2

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Answer 2

Answer:  The required common ratio of the G.P. is 2.

Step-by-step explanation:  Given that the pth, 2pth and 4pth terms of an AP are in G.P.

We are to find the common ratio of the G.P.

We know that the nth term of an AP with first term a and common difference d is given by

$a_n=a+(n-1)d.$

So, we have

$a_p=a+(p-1)d,\\\\a_{2p}=a+(2p-1)d,\\\\a_{4p}=a+(4p-1)d.$

According to the given information, the pth, 2pth and 4pth terms are in G.P., so we must get

$a_{2p}^2=a_p\times a_{4p}\\\\\Rightarrow (a+(2p-1)d)^2=(a+(p-1)d)(a+(4p-1)d)\\\\\Rightarrow a^2+2ad(2p-1)+(2p-1)^2d^2=a^2+ad(4p-1)+ad(p-1)+(p-1)(4p-1)d^2\\\\\Rightarrow adp=d^2p\\\\\Rightarrow a=d.$

So, the pth, 2pth and 4pth terms of the GP are

$a_p=d+(p-1)d=pd,\\\\a_{2p}=d+(2p-1)d=2pd,\\\\a_{4p}=d+(4p-1)d=4pd.$

Therefore, the common ratio of the G.P. is

$r=\dfrac{a_{2p}}{a_p}=\dfrac{a_{4p}}{a_{2p}}=2.$

Thus, the required common ratio of the G.P. is 2.