#### Question

Given a geometric sequence with the common ratio q (q>1) and the number of terms m, removing one of the terms from the sequence will result in a new sequence. Find the values of m and q if the new sequence is an arithmetic sequence.

Collected in the board: Arithmetic sequence

Steven Zheng posted 6 days ago

#### Answer

First it comes to my mind that it is not true for the new sequence to contain 3 conservative terms of the original series. To prove the hypothesis, let assume the statement is true.

a_k, a_{k+1},a_{k+2} satisfies the criteria of an arithmetic sequence. Then we get the following equation.

a_k+a_{k+2}=2a_{k+1}

Since they also form a geometric sequence,

applying the formula for the general terms for the above equation gives

q^{k-1}+q^{k+1}=2q^k

q^k-q^{k-1}=q^{k+1}-q^k

q^{k-1}(q-1)=q^k(q-1)

(q-1)(q^{k-1}-q^k)=0

q^{k-1}(q-1)^2=0

Solving the equation gives q=1, which contradicts with the given condition, q>1. Now we have approved our hypothesis is true. So the number of the terms of the original sequence is either 4 or 5.

When m=4, the sequence looks like

a_1,a_2,a_3, a_4

Taking out either a_2 or a_3 (e.g., a_2), the remaining terms form an arithmetic sequence. That is

2a_3=a_1+a_4

Applying the formula for general terms for an arithmetic sequence,

2a_1q^2=a_1+a_1q^3

q^3-2q^2+1=0

q^3-2q^2+2-1=0

q^3-1-2(q^2-1)=0

(q-1)(q^2+q+1)-(q-1)(2q+2)=0

(q-1) (q^2-q-1)=0

So we get the solutions

q=1 (canceled )

q=\dfrac{1+\sqrt{5} }{2}

And

q=\dfrac{1-\sqrt{5} }{2} (canceled since q>1)

When m=5, the sequence looks like

a_1,a_2,a_3, a_4,a_5

Only a_3 is allowed to take out to avoid 3 conservative terms in the new sequence.

Since the new sequence a_1,a_2,a_4,a_5 is an arithmetic sequence,

a_1+a_4=a_2+a_5

Since they also satisfy the criteria of a geometric sequence,

a_1+a_1q^3=a_1q+a_1q^4

1-q+q^3-q^4=0

(1-q)(1+q^3)=0

q=\pm 1 (canceled )

In summary, the value of m is 4, q=\dfrac{1+\sqrt{5} }{2}

Steven Zheng posted 5 days ago

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