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Someone can explain me the question.. I think so there is a flaw in the question
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In a certain sequence, when you subtract the Mth element from the (M-1)th element, you get twice the Mth element (M is any positive integer). What is the fourth element of this sequence?

(1) The first element of the sequence is 1.
(2) The third element of the sequence is 1/9.

Solution



b1 - b2 = 2b2
b1 = 3b2
b2/b1 = 1/3 => Common Ratio

Need: Any term in sequence

1) b1 = 1
b4 = 1 * (1/3)^3 = 1/27 => Sufficient


2) b3 = 1/3
b4 = (1/9) * (1/3) = 1/27 => Sufficient

ANSWER: D
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Hi Bunuel / ScottTargetTestPrep / KarishmaB

Can you pls help me understand this better. If the question states that M is any positive INTEGER, how can any element in the sequence be a fraction? - which is what we'll get if we have to go by st. 2 and also if we have to just use some numbers for Mth element and (M-1)th element.

Thanks!
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Suruchim12
Hi Bunuel / ScottTargetTestPrep / KarishmaB

Can you pls help me understand this better. If the question states that M is any positive INTEGER, how can any element in the sequence be a fraction? - which is what we'll get if we have to go by st. 2 and also if we have to just use some numbers for Mth element and (M-1)th element.

Thanks!

M is not the actual value in the list. When we say the Mth element, we mean say the 1st element or 2nd element or 3rd element etc. We cannot have a 0th element or 2.1th element and that is why we are given that M must be a positive integer. The actual element in the list can take any fraction or negative value.

Given: "when you subtract the Mth element from the (M-1)th element, you get twice the Mth element"

Take an example: This means that when we subtract the 2nd element from the 1st element, we get twice the 2nd element.

Let's write these terms of the sequence in our standard form, say using \(t_m\) as a generic term.

When m = 2
We are given that:
\(t_1- t_2 = 2 * t_2\)
So,
\(t_2 = \frac{t_1}{3} \)
\(t_3 = \frac{t_2}{3} \)

and so on.. Hence this is a Geometric Progression with common ratio as 1/3.
Now when we are given any term, we can find all other terms (because it gives us the first term a)

Answer (D)

Check out this post on sequences to understand how to deal with them: https://anaprep.com/algebra-introducing-sequences/
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