Answer :
To find the general term of an arithmetic sequence where the sum of the first [tex]\( n \)[/tex] terms is given by [tex]\( S_n = \frac{n}{4}(3n - 1) \)[/tex], we need to follow a systematic approach.
1. Recall the formula for the sum of the first [tex]\( n \)[/tex] terms of an arithmetic sequence:
[tex]\[ S_n = \frac{n}{2}(a_1 + a_n) \][/tex]
where [tex]\( a_1 \)[/tex] is the first term and [tex]\( a_n \)[/tex] is the [tex]\( n \)[/tex]-th term of the sequence.
2. Given the sum [tex]\( S_n = \frac{n}{4}(3n - 1) \)[/tex], equate this to the sum formula:
[tex]\[ \frac{n}{4}(3n - 1) = \frac{n}{2}(a_1 + a_n) \][/tex]
3. Simplify the equation:
[tex]\[ \frac{3n^2 - n}{4} = \frac{n}{2}(a_1 + a_n) \][/tex]
Multiply both sides by 4 to clear the fraction:
[tex]\[ 3n^2 - n = 2n(a_1 + a_n) \][/tex]
Divide both sides by [tex]\( n \)[/tex]:
[tex]\[ 3n - 1 = 2(a_1 + a_n) \][/tex]
4. Express [tex]\( a_n \)[/tex] as:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
where [tex]\( d \)[/tex] is the common difference.
5. Substitute [tex]\( a_n \)[/tex] in the simplified equation:
[tex]\[ 3n - 1 = 2\left(a_1 + (a_1 + (n - 1)d)\right) \][/tex]
Simplify the right side:
[tex]\[ 3n - 1 = 2(2a_1 + (n - 1)d) \][/tex]
6. Distribute and solve for [tex]\( a_1 \)[/tex] and [tex]\( d \)[/tex]:
[tex]\[ 3n - 1 = 4a_1 + 2nd - 2d \][/tex]
Group the terms involving [tex]\( n \)[/tex]:
[tex]\[ 3n - 1 = 4a_1 + 2nd - 2d \][/tex]
[tex]\[ 3n - 1 = 4a_1 + 2nd - 2d \][/tex]
7. Compare coefficients on both sides:
For the coefficient of [tex]\( n \)[/tex]:
[tex]\[ 3 = 2d \implies d = \frac{3}{2} \][/tex]
For the constant term:
[tex]\[ -1 = 4a_1 - 2d \][/tex]
Substitute [tex]\( d = \frac{3}{2} \)[/tex]:
[tex]\[ -1 = 4a_1 - 2\left(\frac{3}{2}\right) \][/tex]
Simplify:
[tex]\[ -1 = 4a_1 - 3 \][/tex]
Solve for [tex]\( a_1 \)[/tex]:
[tex]\[ 2 = 4a_1 \implies a_1 = \frac{1}{2} \][/tex]
8. Construct the general term [tex]\( a_n \)[/tex]:
Using [tex]\( a_1 = \frac{1}{2} \)[/tex] and [tex]\( d = \frac{3}{2} \)[/tex]:
[tex]\[ a_n = a_1 + (n - 1)d \][/tex]
[tex]\[ a_n = \frac{1}{2} + (n - 1)\frac{3}{2} \][/tex]
Simplify:
[tex]\[ a_n = \frac{1}{2} + \frac{3}{2}n - \frac{3}{2} \][/tex]
[tex]\[ a_n = \frac{3}{2}n - 1 \][/tex]
Therefore, the general term of the sequence is:
[tex]\[ a_n = \frac{3}{2}n - 1 \][/tex]
1. Recall the formula for the sum of the first [tex]\( n \)[/tex] terms of an arithmetic sequence:
[tex]\[ S_n = \frac{n}{2}(a_1 + a_n) \][/tex]
where [tex]\( a_1 \)[/tex] is the first term and [tex]\( a_n \)[/tex] is the [tex]\( n \)[/tex]-th term of the sequence.
2. Given the sum [tex]\( S_n = \frac{n}{4}(3n - 1) \)[/tex], equate this to the sum formula:
[tex]\[ \frac{n}{4}(3n - 1) = \frac{n}{2}(a_1 + a_n) \][/tex]
3. Simplify the equation:
[tex]\[ \frac{3n^2 - n}{4} = \frac{n}{2}(a_1 + a_n) \][/tex]
Multiply both sides by 4 to clear the fraction:
[tex]\[ 3n^2 - n = 2n(a_1 + a_n) \][/tex]
Divide both sides by [tex]\( n \)[/tex]:
[tex]\[ 3n - 1 = 2(a_1 + a_n) \][/tex]
4. Express [tex]\( a_n \)[/tex] as:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
where [tex]\( d \)[/tex] is the common difference.
5. Substitute [tex]\( a_n \)[/tex] in the simplified equation:
[tex]\[ 3n - 1 = 2\left(a_1 + (a_1 + (n - 1)d)\right) \][/tex]
Simplify the right side:
[tex]\[ 3n - 1 = 2(2a_1 + (n - 1)d) \][/tex]
6. Distribute and solve for [tex]\( a_1 \)[/tex] and [tex]\( d \)[/tex]:
[tex]\[ 3n - 1 = 4a_1 + 2nd - 2d \][/tex]
Group the terms involving [tex]\( n \)[/tex]:
[tex]\[ 3n - 1 = 4a_1 + 2nd - 2d \][/tex]
[tex]\[ 3n - 1 = 4a_1 + 2nd - 2d \][/tex]
7. Compare coefficients on both sides:
For the coefficient of [tex]\( n \)[/tex]:
[tex]\[ 3 = 2d \implies d = \frac{3}{2} \][/tex]
For the constant term:
[tex]\[ -1 = 4a_1 - 2d \][/tex]
Substitute [tex]\( d = \frac{3}{2} \)[/tex]:
[tex]\[ -1 = 4a_1 - 2\left(\frac{3}{2}\right) \][/tex]
Simplify:
[tex]\[ -1 = 4a_1 - 3 \][/tex]
Solve for [tex]\( a_1 \)[/tex]:
[tex]\[ 2 = 4a_1 \implies a_1 = \frac{1}{2} \][/tex]
8. Construct the general term [tex]\( a_n \)[/tex]:
Using [tex]\( a_1 = \frac{1}{2} \)[/tex] and [tex]\( d = \frac{3}{2} \)[/tex]:
[tex]\[ a_n = a_1 + (n - 1)d \][/tex]
[tex]\[ a_n = \frac{1}{2} + (n - 1)\frac{3}{2} \][/tex]
Simplify:
[tex]\[ a_n = \frac{1}{2} + \frac{3}{2}n - \frac{3}{2} \][/tex]
[tex]\[ a_n = \frac{3}{2}n - 1 \][/tex]
Therefore, the general term of the sequence is:
[tex]\[ a_n = \frac{3}{2}n - 1 \][/tex]
