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26 Dec 2023, 06:33
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\(a_{1}\): 4
\(d\): 6
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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58% (01:58) correct
42%
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In an arithmetic sequence, the sum, \(S_{n}\) of the first \(n\) terms of the sequence equals \(n + 3n^2\) for all \(n\).
In the table, select a value for the first term, \(a_{1}\), of the sequence, and a value for the difference, \(d\), between successive terms of the sequence.
In the table, select a value for the first term, \(a_{1}\), of the sequence, and a value for the difference, \(d\), between successive terms of the sequence.
| \(a_{1}\) | \(d\) | |
| 2 | ||
| 4 | ||
| 6 | ||
| 8 | ||
| 10 | ||
| 12 |
ShowHide Answer
Official Answer
\(a_{1}\): 4
\(d\): 6
26 Dec 2023, 06:57
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Sajjad1994
\(S_n\) is the sum of the first n terms.
Substitute n=1: \(S_1=n + 3n^2=1+3*1^2=4\), but \(S_1\) is nothing but the first term. So, \(a_1=4\)
Substitute n=2: \(S_2=n + 3n^2=2+3*2^2=14\), but \(S_2\) is nothing but the sum of first two terms => \(S_2=a_1+a_2......14=4+a_2.......a_2=10\). So, \(d=a_2-a_1=10-4=6\)
General Discussion
24 Aug 2024, 23:41
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Sn is the sum of the first n terms.
Given, sequence is an Arithmentic Sequence. And sum of first n terms of that sequence is: \(n + 3n^2\) ---> Eq 1
Formula for for sum of first n terms of an Arithmetic Sequence is: \( Sn= n/2* [ 2a + ( n − 1 )d ]\). ----> Eq 2
Equate 1 and 2.
\( n + 3n^2 = n /2* [ 2a + ( n − 1 )d ]\)
Expand the above -
\(2n + 6n^2= (2a-d)n + d(n^2)\)
Compare n^2 terms on LHS and RHS -
\( 6n^2 = d(n^2) \)
d=6
Compare n terms on LHS and RHS -
\((2a-d)n=2n\)
\(2a-d=2\)
Substituting d=6;
\(2a=8\)
a=4
Given, sequence is an Arithmentic Sequence. And sum of first n terms of that sequence is: \(n + 3n^2\) ---> Eq 1
Formula for for sum of first n terms of an Arithmetic Sequence is: \( Sn= n/2* [ 2a + ( n − 1 )d ]\). ----> Eq 2
Equate 1 and 2.
\( n + 3n^2 = n /2* [ 2a + ( n − 1 )d ]\)
Expand the above -
\(2n + 6n^2= (2a-d)n + d(n^2)\)
Compare n^2 terms on LHS and RHS -
\( 6n^2 = d(n^2) \)
d=6
Compare n terms on LHS and RHS -
\((2a-d)n=2n\)
\(2a-d=2\)
Substituting d=6;
\(2a=8\)
a=4
