Given that the first term of an arithmetic sequence is a and the common difference is d. The 3rd, 4th, and 7th term of this arithmetic sequence are the first three terms of a geometric sequence. Hence, show that $a=\frac{-3}{2}d$

So for the arithmetic sequence the terms would be, $a, a+d, a+2d, a+3d ...$

And for the Geometric sequence it would be, $a+2d, a+3d, a+6d$

So I tried solving for d using the ratio between the geometric terms. I didn't get the correct answer, can someone point me in the correct direction?


Suppose that none of 3rd, 4th or 7th terms in the sequence are zero.

We get then that the ratio of consecutive terms in the geometric sequence must be equal, that is:


By cross multiplying we get:

$$a^2+8ad+12d^2 = a^2+6ad+9d^2$$

By subtracting some terms and simplifying:

$$2ad = -3d^2$$

Now, assuming that $d\neq 0$ we can simplify a bit further and get:



Recall the definition of a geometric sequence. The ratio between $u_{n+1}$ and $u_n$ is kept constant.

$$u_{n-1}, u_n, u_{n+1}$$

$$r = \frac{u_{n+1}}{u_n} = \frac{u_n}{u_{n-1}}$$

$$\implies (u_n)^2 = u_{n-1}\cdot u_{n+1}$$

The square of any $n^{th}$ term is equal to the product of the $(n-1)^{th}$ and $(n+1)^{th}$ terms (the terms before and after it).

Apply the same logic to the question. You already wrote the first $3$ terms of the geometric sequence.

$$a+2d, a+3d, a+6d$$

$$(a+3d)^2 = (a+2d)(a+6d)$$

From here, all that you need to do is expand, simplify, and then solve for $a$.

$$a^2+6ad+9d^2 = a^2+8ad+12d^2$$ $$0 = 3d^2+2ad$$ $$0 = d(3d+2a) \implies 0 = 3d+2a$$ $$2a = -3d \implies a = -\frac{3}{2}$$

  • $\begingroup$ You are using the letter $d$ in two different ways here. $\endgroup$ Nov 8 '18 at 22:37
  • $\begingroup$ @N. F. Taussig The part where I wrote $d$ for ratio? $\endgroup$
    – KM101
    Nov 8 '18 at 22:38
  • $\begingroup$ Yes, you used $d$ both as a common ratio and as the common difference of the arithmetic sequence. $\endgroup$ Nov 8 '18 at 22:39
  • 1
    $\begingroup$ Thanks for the catch, I corrected it. I meant to use $r$ for ratio. $\endgroup$
    – KM101
    Nov 8 '18 at 22:40

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