My ‘attempt’ is pretty disastrous but I haven’t found a better way to explain what I’m doing. Please bear with me.

Q.The 1st, 3rd and 6th terms of an arithmetic sequence are also in geometric sequence. Find the common ratio (r) of the geometric sequence.

From text book, r = 3/ 2

A.:

If Un = a + (n - 1)d, assume d = 1, 2 and 3 respectively. This will result in the following sequences:

Attempt:

If d = 1: Un-1 = 1, 2, 3, 4, 5, 6

If d = 2: Un-1 = 2, 4, 6, 8, 10, 12

If d = 3: Un-1 = 3, 6, 9 12, 15, 18

From the question statement, we know that the 1st, 3rd and 6th numbers of the arithmetic sequence are in order for the geometric sequence.

In sequence Un-1, where d = 1

1, 3, 6 = U1, U2 and U3 of the geometric sequence respectively.

Similarly, where d = 2, we have:

2, 6, 12

And when d = 3:

3, 9, 18

Note in each case: U3/ U2 = 2 and U2/ U1 = 3, e.g. in the sequence where d = 3:

18/ 9 = 2 & 9/ 3 = 2

Therefore, r = 3/2, i.e (U2/ U1)/ (U3/ U2)…?

However,…

Geometric sequence: Un = ar^n-1

Going by geometric sequence layout, U1, U2, U3 = a, ar, ar^2, respectively.

Which means that in, say Un-1, d = 3, the sequence 3, 9, 18 should really be 3, 9, 27, i.e we say r = 3 and so: U1 = 3, U2 = 3.3, U3 = 3.3^2

Can anyone help me confirm if I‘m right, or completely off the mark? Thank you.