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Bunuel
In a certain sequence, the term an is defined by the formula \(a_n = a_{n – 1} + 10\) for each integer n ≥ 2. What is the positive difference between \(a_{10}\) and \(a_{15}\)?

An = An-1 + 10

A15 = A14 + 10
A15 = A13 + 10 + 10
A15 = A12 + 10 + 10 + 10
A15 = A11 + 10 + 10 + 10 + 10
A15 = A10 + 10 + 10 + 10 + 10 + 10
A15 - A10 = 10 + 10 + 10 + 10 + 10
A15 - A10 = 50

Hence, D.



Thanks for the explanation, your approach absolutely makes sense to me.

Can somebody verify that the following approach is valid as well:

I know n is defined as 2 or bigger but plugging in 1 for n leaves as with a1 = a0 +10 and a0 has to be 0 as there is no 0th/st sequence.

Hence a1=10,a2=20,a3=30,a4=40,a5=50 and therefore a10 = 100 and a15 = 150

100-150 = -50 and since we are looking for the positive difference D is the final answer choice.


Thanks for your help in advance and kind regards,
Max
maxmayr

Quote:
In a certain sequence, the term an is defined by the formula \(a_n = a_{n – 1} + 10\) for each integer n ≥ 2.
Though you are getting a correct answer,you are assuming \(a_n = a_{n – 1} + 10\) is valid for n<2 also.
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Quote:
In a certain sequence, the term an is defined by the formula \(a_n = a_{n – 1} + 10\) for each integer n ≥ 2.
Though you are getting a correct answer,you are assuming \(a_n = a_{n – 1} + 10\) is valid for n<2 also.

Yes, absolutely true, but then I have to ask where do we get information to conclude that:

if A15 = A14 + 10, then A15 = A13 + 10 + 10

We substitute A14 automatically with 10.

What do I miss out here?

Kind regards,
Max
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Quote:
In a certain sequence, the term an is defined by the formula \(a_n = a_{n – 1} + 10\) for each integer n ≥ 2.
Though you are getting a correct answer,you are assuming \(a_n = a_{n – 1} + 10\) is valid for n<2 also.

Yes, absolutely true, but then I have to ask where do we get information to conclude that:

if A15 = A14 + 10, then A15 = A13 + 10 + 10

We substitute A14 automatically with 10.

What do I miss out here?

Kind regards,
Max

\(A_n=A_{n-1}+10\)
\(A_{15}=A_{14} +10\) ...(1)
\(A_{14}=A_{13} +10\) ...(2)
\(A_{13}=A_{12} +10\) ...(3)
\(A_{12}=A_{11} +10\) ...(4)
\(A_{11}=A_{10} +10\) ...(5)
Substituting (5) into (4)
We get
\(A_{12}= A_{10} +10+10\) ...(6)
Substituting (6) into (3)
\(A_{13}=A_{10} +10+10 +10\) ...(7)
Substituting (7) into (2)
\(A_{14}=A_{10} +10+10 +10 +10\) ...(8)
Substituting (8) into (1)
\(A_{15}=A_{10} +10+10 +10 +10 +10\)
\(A_{15}=A_{10} +50\)
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