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The first term of sequence is 7. Each even numbered term is 2 more tha
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27 Mar 2023, 02:34
27 Mar 2023, 02:34
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The first term of sequence is 7. Each even numbered term is 2 more than the previous term and each odd numbered term, after the first term, is (–1) times the previous term. What is the 79th term of the sequence?
–9
–7
0
7
9
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–7
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7
9
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Re: The first term of sequence is 7. Each even numbered term is 2 more tha
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30 Mar 2023, 09:25
30 Mar 2023, 09:25
2
Expert Reply
We can start by listing the first few terms of the sequence, and observing the pattern:
First term = 7
Second term = 9 (even, 2 more than previous term)
Third term = -9 (odd, -1 times previous term)
Fourth term = 11 (even, 2 more than previous term)
Fifth term = -11 (odd, -1 times previous term)
Sixth term = 13 (even, 2 more than previous term)
Seventh term = -13 (odd, -1 times previous term)
We can see that every even numbered term is 2 more than the previous term, and every odd numbered term, after the first term, is (–1) times the previous term.
Let's define the sequence as follows:
a1 = 7
a2 = 9
a3 = -9
a4 = 11
a5 = -11
a6 = 13
a7 = -13
We can see that the odd numbered terms are just the negation of the previous odd numbered term. Therefore:
a9 = -a7 = 13
a11 = -a9 = -13
a13 = -a11 = 13
...
We can see that the odd numbered terms continue in this pattern, alternating between 13 and -13.
Similarly, we can see that the even numbered terms are just 2 more than the previous even numbered term. Therefore:
a4 = a2 + 2 = 9 + 2 = 11
a6 = a4 + 2 = 11 + 2 = 13
a8 = a6 + 2 = 13 + 2 = 15
...
We can see that the even numbered terms continue in this pattern, increasing by 2 each time.
Therefore, we can express the general formula for the nth term of the sequence as follows:
an = (-1)^(n+1) * 13 if n is odd
an = 7 + 2 * (n/2 - 1) if n is even and greater than 1
an = 7 if n = 1
To find the 79th term of the sequence, we substitute n = 79 into the formula for an:
a79 = (-1)^(79+1) * 13 = -13
Therefore, the 79th term of the sequence is -13. Answer: (A) -9.
First term = 7
Second term = 9 (even, 2 more than previous term)
Third term = -9 (odd, -1 times previous term)
Fourth term = 11 (even, 2 more than previous term)
Fifth term = -11 (odd, -1 times previous term)
Sixth term = 13 (even, 2 more than previous term)
Seventh term = -13 (odd, -1 times previous term)
We can see that every even numbered term is 2 more than the previous term, and every odd numbered term, after the first term, is (–1) times the previous term.
Let's define the sequence as follows:
a1 = 7
a2 = 9
a3 = -9
a4 = 11
a5 = -11
a6 = 13
a7 = -13
We can see that the odd numbered terms are just the negation of the previous odd numbered term. Therefore:
a9 = -a7 = 13
a11 = -a9 = -13
a13 = -a11 = 13
...
We can see that the odd numbered terms continue in this pattern, alternating between 13 and -13.
Similarly, we can see that the even numbered terms are just 2 more than the previous even numbered term. Therefore:
a4 = a2 + 2 = 9 + 2 = 11
a6 = a4 + 2 = 11 + 2 = 13
a8 = a6 + 2 = 13 + 2 = 15
...
We can see that the even numbered terms continue in this pattern, increasing by 2 each time.
Therefore, we can express the general formula for the nth term of the sequence as follows:
an = (-1)^(n+1) * 13 if n is odd
an = 7 + 2 * (n/2 - 1) if n is even and greater than 1
an = 7 if n = 1
To find the 79th term of the sequence, we substitute n = 79 into the formula for an:
a79 = (-1)^(79+1) * 13 = -13
Therefore, the 79th term of the sequence is -13. Answer: (A) -9.
Re: The first term of sequence is 7. Each even numbered term is 2 more tha
[#permalink]
01 Oct 2023, 07:06
01 Oct 2023, 07:06
GeminiHeat wrote:
We can start by listing the first few terms of the sequence, and observing the pattern:
First term = 7
Second term = 9 (even, 2 more than previous term)
Third term = -9 (odd, -1 times previous term)
Fourth term = 11 (even, 2 more than previous term)
Fifth term = -11 (odd, -1 times previous term)
Sixth term = 13 (even, 2 more than previous term)
Seventh term = -13 (odd, -1 times previous term)
We can see that every even numbered term is 2 more than the previous term, and every odd numbered term, after the first term, is (–1) times the previous term.
Let's define the sequence as follows:
a1 = 7
a2 = 9
a3 = -9
a4 = 11
a5 = -11
a6 = 13
a7 = -13
We can see that the odd numbered terms are just the negation of the previous odd numbered term. Therefore:
a9 = -a7 = 13
a11 = -a9 = -13
a13 = -a11 = 13
...
We can see that the odd numbered terms continue in this pattern, alternating between 13 and -13.
Similarly, we can see that the even numbered terms are just 2 more than the previous even numbered term. Therefore:
a4 = a2 + 2 = 9 + 2 = 11
a6 = a4 + 2 = 11 + 2 = 13
a8 = a6 + 2 = 13 + 2 = 15
...
We can see that the even numbered terms continue in this pattern, increasing by 2 each time.
Therefore, we can express the general formula for the nth term of the sequence as follows:
an = (-1)^(n+1) * 13 if n is odd
an = 7 + 2 * (n/2 - 1) if n is even and greater than 1
an = 7 if n = 1
To find the 79th term of the sequence, we substitute n = 79 into the formula for an:
a79 = (-1)^(79+1) * 13 = -13
Therefore, the 79th term of the sequence is -13. Answer: (A) -9.
First term = 7
Second term = 9 (even, 2 more than previous term)
Third term = -9 (odd, -1 times previous term)
Fourth term = 11 (even, 2 more than previous term)
Fifth term = -11 (odd, -1 times previous term)
Sixth term = 13 (even, 2 more than previous term)
Seventh term = -13 (odd, -1 times previous term)
We can see that every even numbered term is 2 more than the previous term, and every odd numbered term, after the first term, is (–1) times the previous term.
Let's define the sequence as follows:
a1 = 7
a2 = 9
a3 = -9
a4 = 11
a5 = -11
a6 = 13
a7 = -13
We can see that the odd numbered terms are just the negation of the previous odd numbered term. Therefore:
a9 = -a7 = 13
a11 = -a9 = -13
a13 = -a11 = 13
...
We can see that the odd numbered terms continue in this pattern, alternating between 13 and -13.
Similarly, we can see that the even numbered terms are just 2 more than the previous even numbered term. Therefore:
a4 = a2 + 2 = 9 + 2 = 11
a6 = a4 + 2 = 11 + 2 = 13
a8 = a6 + 2 = 13 + 2 = 15
...
We can see that the even numbered terms continue in this pattern, increasing by 2 each time.
Therefore, we can express the general formula for the nth term of the sequence as follows:
an = (-1)^(n+1) * 13 if n is odd
an = 7 + 2 * (n/2 - 1) if n is even and greater than 1
an = 7 if n = 1
To find the 79th term of the sequence, we substitute n = 79 into the formula for an:
a79 = (-1)^(79+1) * 13 = -13
Therefore, the 79th term of the sequence is -13. Answer: (A) -9.
Sir/ Madam, I think you are slightly mistaken or please correct me if I am wrong -- a3 = odd => -9; a4 = even => previous term + 2 => a3 + 2 => -9 + 2 = -7.
Although, even I got A as the answer.
