Question Video: Identifying Arithmetic Sequences given Their 𝑛th Term | Nagwa Question Video: Identifying Arithmetic Sequences given Their 𝑛th Term | Nagwa

# Question Video: Identifying Arithmetic Sequences given Their πth Term Mathematics • Second Year of Secondary School

## Join Nagwa Classes

Which of the following is an arithmetic sequence? [A] π_(π) = β(8)^(π) [B] π_(π) = β9πΒ² β 8π + 1 [C] π_(π) = πΒ² β 1 [D] π_(π) = β(π β 7) [E] π_(π) = β9π + 1

07:48

### Video Transcript

Which of the following is an arithmetic sequence? Is it option (A) π sub π is equal to negative eight all raised to the πth power? Is it option (B) π sub π is equal to negative nine π squared minus eight π plus one? Is it option (C) π sub π is equal to π squared minus one? Option (D) π sub π is equal to the square root of π minus seven. Or is it option (E) π sub π is equal to negative nine π plus one?

In this question, weβre given five possible sequences, and we need to determine which of these sequences is an arithmetic sequence. To answer this question, letβs start by recalling what we mean by an arithmetic sequence. This is a sequence where the difference between any two consecutive terms is constant. So we need to determine which of the five sequences given has this property. Before we do this, letβs remind ourselves what we mean by the notation π sub π. In this notation, the value of π is the term number of our sequence. So if we substitute π is equal to one into this expression, we get the first term of the sequence. So in our first option, π sub one is equal to negative eight to the first power, which simplifies to give us negative eight. So negative eight is the first term in the sequence.

We can then do the same to find the second term in our sequence. We substitute π is equal to two into this expression and simplify. The second term in the sequence is 64. We can do exactly the same to find the third term in the sequence. We substitute π is equal to three into this expression to get π sub three is equal to negative eight cubed. And π cubed is equal to 512. So the third term in the sequence is negative 512. And remember, in an arithmetic sequence, the difference between any two consecutive terms is constant. Therefore, for the sequence to be an arithmetic sequence, the difference between the first and second term of the sequence must be equal to the difference between the second and third term of the sequence.

We can calculate both of these values. First, the difference between the third and second term of the sequence is negative 512 minus 64, which we can calculate is negative 576. And we can then calculate the difference between the second and first term of the sequence. Thatβs 64 minus eight, which if we calculate is equal to 72. These values are not equal to each other, so the difference between consecutive terms does not remain constant. Therefore, option (A) is not an arithmetic sequence.

Letβs clear some space and then do the same for option (B). We find the first term of the sequence by substituting one into the expression. π sub one is equal to negative nine times one squared minus eight times one plus one which, if we simplify and evaluate, we see itβs equal to negative 16. We have π sub two is equal to negative nine times two squared minus eight times two plus one, which we can then evaluate to see that π sub two is equal to negative 51. And we can do the same for the third term in the sequence. We see the third term of the sequence is negative 104.

And now, weβre ready to check if the difference between the third and second term of the sequence and the second and first term of the sequence is constant. First, the difference between the third and second term of the sequence is negative 104 minus negative 51, which we can evaluate is equal to negative 53. And we can do the same to find the difference between the second and first term. Itβs negative 51 minus negative 16, which is equal to negative 35. And we can see the difference between these consecutive terms is not equal. Therefore, the difference between consecutive terms in the sequence does not remain constant, so it is not an arithmetic sequence.

We can do exactly the same thing for option (C). The first term in the sequence is one squared minus one, which is equal to zero. The second term in the sequence is two squared minus one, which is equal to three. And the third term in the sequence is three squared minus one, which is equal to eight. We can then once again check if the difference between consecutive terms remains constant. First, the difference between the third and second term is eight minus three, which we can calculate is equal to five. And the difference between the second and first term of the sequence is three minus zero, which is equal to three. Therefore, the difference between the second and first term of the sequence and the third and second term of the sequence is not equal. The difference between consecutive terms does not remain constant, so (C) is not an arithmetic sequence.

So weβll clear some space and move on to option (D). We can find the first term in option (D) by substituting π is equal to one. Itβs equal to the square root of one minus seven, which is the square root of negative six. And then, by using our laws of exponents and our knowledge of imaginary numbers, this simplifies to give us root six times π, which is an imaginary number. So we now need to ask the question, are we allowed to have arithmetic sequences which contain imaginary numbers? And the answer is, it depends. Some people demand that their sequences only contain real numbers. If this is the case, then we can conclude option (D) is incorrect because root six π is not a real number.

However, it is possible to define an arithmetic sequence by using complex numbers in exactly the same way. So weβll just check if this is an arithmetic sequence in exactly the same way. We find the second and third term of this sequence by substituting π is equal to two and π is equal to three into this expression, respectively.

The second term of the sequence is root five π and the third term of the sequence is two π. We then check if this is an arithmetic sequence in the same way. We need to check that the difference between consecutive terms remains constant. The difference between the third and second term is two π minus root five π, which we can simplify to two minus root five all multiplied by π. And we can calculate the difference between the second and first term. We get root five π minus root six π, which simplifies to give us root five minus root six all multiplied by π.

And we can see that these are not equal because the coefficients of π in both of these two numbers are not the same. The imaginary parts of the two numbers are different. And therefore, the difference between consecutive terms in the sequence does not remain constant. (D) is not an arithmetic sequence.

Finally, letβs check option (E). We find the first term of the sequence by substituting π is equal to one. The first term of the sequence is negative eight. We substitute π is equal to two to find that the second term of the sequence is negative 17. And in the same way, we find the third term of the sequence is negative 26. Now, we need to check if the difference between these consecutive terms remains constant. The difference between the third and second term is negative 26 minus negative 17, which we can calculate is negative nine. And in the same way, the difference between the second and first term is negative 17 minus negative eight, which is also equal to negative nine.

We might want to just conclude that this means option (E) is an arithmetic sequence. However, we need to be careful because we need to check the difference between any two consecutive terms remains constant. And thereβs a few different ways of doing this. One way is to determine the difference between the term π sub π plus one and the term π sub π. We can find an expression for π sub π plus one by substituting π plus one into our expression. We get negative nine times π plus one plus one. Distributing and simplifying this expression, we get negative nine π minus eight.

And we already know that π sub π is equal to negative nine plus one. We can now determine the difference between the π plus oneth term of our sequence and the πth term in our sequence. The difference between these two terms is negative nine π minus eight minus negative nine π plus one. Distributing the negative over our parentheses gives us negative nine π minus eight plus nine π minus one. And if we simplify this expression, weβre left with negative nine, which is a constant value. Therefore, for any positive integer value of π we choose, the difference between consecutive terms remains at constant value of negative nine. So the sequence in option (E) is an arithmetic sequence.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions