The 10th, 4th and 1st term of an A. P are three consecutive find the common ratio of the G. P and the sum of the 6th term, taking the 1st term to be 4

S_6 = 252

Step-by-step explanation:

We are told that the 10th, 4th and 1st term of an A.P are three consecutive terms of a G.P.

Now,formula for nth term of an AP is;

a_n = a + (n - 1)d

Thus;

a_10 = a + (10 - 1)d

a_10 = a + 9d

Also;

a_4 = a + (4 - 1)d

a_4 = a + 3d

First term is a.

Thus, since they are consecutive terms of a G.P, it means that;

(a + 9d)/(a + 3d) = (a + 3d)/a

Cross multiply to get;

a(a + 9d) = (a + 3d)(a + 3d)

a² + 9ad = a² + 6ad + 9d²

a² will cancel out to give;

9ad - 6ad = 9d²

3ad = 9d²

Divide both sides by 3d to get;

a = 3d

We are told that the first term is 4.

Thus, 4 = 3d

d = 4/3

We saw earlier that ratio of the GP is (a + 3d)/a

Thus; r = (4 + 3(4/3))/4 = 8/4 = 2

Sum of n terms of a G.P is given by;

S_n = a(rⁿ - 1)/(r - 1)

S_6 = 4(2^(6) - 1)/(2 - 1)

S_6 = 252

table 1

step-by-step explanation:

1-16

2-32

4-64

8-128

all multiples of 16