Different methods to solving a sequence problem
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This is a question from Manhattan Prep GRE book, and here is the question.
If each number in a sequence is three more than the previous number, and the sixth number is $32$, what is the 100th number?
what is the sequence rule to this problem. While the book finds the solution differently. I would like to know what the sequence rule is.
This is solution Manhattan Prep provided.
$100 - 6 = 94$ terms. $94 *3 = 282$.
$32 + 282 = 314$ . One hundredth term $ = 314$ .
sequences-and-series
asked Dec 22 '18 at 18:29
muhe31muhe31
85
$endgroup$
add a comment |
$begingroup$
This is a question from Manhattan Prep GRE book, and here is the question.
If each number in a sequence is three more than the previous number, and the sixth number is $32$, what is the 100th number?
what is the sequence rule to this problem. While the book finds the solution differently. I would like to know what the sequence rule is.
This is solution Manhattan Prep provided.
$100 - 6 = 94$ terms. $94 *3 = 282$.
$32 + 282 = 314$ . One hundredth term $ = 314$ .
sequences-and-series
asked Dec 22 '18 at 18:29
muhe31muhe31
85
$endgroup$
2
$begingroup$
The sequence is $a,a+3, a+6, ldots , a+3(n-1), ldots$. The sixth term is $a+15$. So $a=17$. Hope you get the idea. This is what is called an arithmetic progression.
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– Anurag A
Dec 22 '18 at 18:31
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Thank you @AnuragA. This helps a lot.
$endgroup$
– muhe31
Dec 22 '18 at 18:35
1
$begingroup$
@AnuragA So to find the 100th term is it a(100th) = 17 + 3(100 - 1) = 314.
$endgroup$
– muhe31
Dec 22 '18 at 18:50
add a comment |
$begingroup$
This is a question from Manhattan Prep GRE book, and here is the question.
If each number in a sequence is three more than the previous number, and the sixth number is $32$, what is the 100th number?
what is the sequence rule to this problem. While the book finds the solution differently. I would like to know what the sequence rule is.
This is solution Manhattan Prep provided.
$100 - 6 = 94$ terms. $94 *3 = 282$.
$32 + 282 = 314$ . One hundredth term $ = 314$ .
sequences-and-series
asked Dec 22 '18 at 18:29
muhe31muhe31
85
$endgroup$
This is a question from Manhattan Prep GRE book, and here is the question.
If each number in a sequence is three more than the previous number, and the sixth number is $32$, what is the 100th number?
what is the sequence rule to this problem. While the book finds the solution differently. I would like to know what the sequence rule is.
This is solution Manhattan Prep provided.
$100 - 6 = 94$ terms. $94 *3 = 282$.
$32 + 282 = 314$ . One hundredth term $ = 314$ .
sequences-and-series
sequences-and-series
asked Dec 22 '18 at 18:29
muhe31muhe31
85
asked Dec 22 '18 at 18:29
muhe31muhe31
85
edited Dec 22 '18 at 18:37
muhe31
asked Dec 22 '18 at 18:29
muhe31muhe31
85
asked Dec 22 '18 at 18:29
muhe31muhe31
85
asked Dec 22 '18 at 18:29
muhe31muhe31
85
85
2
$begingroup$
The sequence is $a,a+3, a+6, ldots , a+3(n-1), ldots$. The sixth term is $a+15$. So $a=17$. Hope you get the idea. This is what is called an arithmetic progression.
$endgroup$
– Anurag A
Dec 22 '18 at 18:31
$begingroup$
Thank you @AnuragA. This helps a lot.
$endgroup$
– muhe31
Dec 22 '18 at 18:35
1
$begingroup$
@AnuragA So to find the 100th term is it a(100th) = 17 + 3(100 - 1) = 314.
$endgroup$
– muhe31
Dec 22 '18 at 18:50
add a comment |
2
$begingroup$
The sequence is $a,a+3, a+6, ldots , a+3(n-1), ldots$. The sixth term is $a+15$. So $a=17$. Hope you get the idea. This is what is called an arithmetic progression.
$endgroup$
– Anurag A
Dec 22 '18 at 18:31
$begingroup$
Thank you @AnuragA. This helps a lot.
$endgroup$
– muhe31
Dec 22 '18 at 18:35
1
$begingroup$
@AnuragA So to find the 100th term is it a(100th) = 17 + 3(100 - 1) = 314.
$endgroup$
– muhe31
Dec 22 '18 at 18:50
2
2
$begingroup$
The sequence is $a,a+3, a+6, ldots , a+3(n-1), ldots$. The sixth term is $a+15$. So $a=17$. Hope you get the idea. This is what is called an arithmetic progression.
$endgroup$
– Anurag A
Dec 22 '18 at 18:31
$begingroup$
The sequence is $a,a+3, a+6, ldots , a+3(n-1), ldots$. The sixth term is $a+15$. So $a=17$. Hope you get the idea. This is what is called an arithmetic progression.
$endgroup$
– Anurag A
Dec 22 '18 at 18:31
$begingroup$
Thank you @AnuragA. This helps a lot.
$endgroup$
– muhe31
Dec 22 '18 at 18:35
$begingroup$
Thank you @AnuragA. This helps a lot.
$endgroup$
– muhe31
Dec 22 '18 at 18:35
1
1
$begingroup$
@AnuragA So to find the 100th term is it a(100th) = 17 + 3(100 - 1) = 314.
$endgroup$
– muhe31
Dec 22 '18 at 18:50
$begingroup$
@AnuragA So to find the 100th term is it a(100th) = 17 + 3(100 - 1) = 314.
$endgroup$
– muhe31
Dec 22 '18 at 18:50
add a comment |
1 Answer
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$begingroup$
hint
$$a_6=32$$
$$a_7=32+3(7-6)$$
$$a_8=32+3(8-6)$$
$$a_{100}=32+3(100-...)$$
answered Dec 22 '18 at 18:36
hamam_Abdallahhamam_Abdallah
38.2k21634
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add a comment |
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1 Answer
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$begingroup$
hint
$$a_6=32$$
$$a_7=32+3(7-6)$$
$$a_8=32+3(8-6)$$
$$a_{100}=32+3(100-...)$$
answered Dec 22 '18 at 18:36
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
add a comment |
$begingroup$
hint
$$a_6=32$$
$$a_7=32+3(7-6)$$
$$a_8=32+3(8-6)$$
$$a_{100}=32+3(100-...)$$
answered Dec 22 '18 at 18:36
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
add a comment |
$begingroup$
hint
$$a_6=32$$
$$a_7=32+3(7-6)$$
$$a_8=32+3(8-6)$$
$$a_{100}=32+3(100-...)$$
answered Dec 22 '18 at 18:36
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
hint
$$a_6=32$$
$$a_7=32+3(7-6)$$
$$a_8=32+3(8-6)$$
$$a_{100}=32+3(100-...)$$
answered Dec 22 '18 at 18:36
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Dec 22 '18 at 18:36
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Dec 22 '18 at 18:36
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Dec 22 '18 at 18:36
hamam_Abdallahhamam_Abdallah
38.2k21634
38.2k21634
add a comment |
add a comment |
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2
$begingroup$
The sequence is $a,a+3, a+6, ldots , a+3(n-1), ldots$. The sixth term is $a+15$. So $a=17$. Hope you get the idea. This is what is called an arithmetic progression.
$endgroup$
– Anurag A
Dec 22 '18 at 18:31
$begingroup$
Thank you @AnuragA. This helps a lot.
$endgroup$
– muhe31
Dec 22 '18 at 18:35
1
$begingroup$
@AnuragA So to find the 100th term is it a(100th) = 17 + 3(100 - 1) = 314.
$endgroup$
– muhe31
Dec 22 '18 at 18:50