Question

Calculate the sum of the first 36 terms of the arithmetic sequence in which a subscript 36 baseline equals 14 and the common difference is d equals Start Fraction 1 over 8 End Fraction.
(1 point)
Responses

425.25
425.25

850.5
850.5

423
423

846
846

Answer 1

To find the sum of the first 36 terms of an arithmetic sequence, we can use the formula for the sum of the first \( n \) terms of an arithmetic sequence:

\[
S_n = \frac{n}{2} \times (a_1 + a_n)
\]

where:
- \( S_n \) is the sum of the first \( n \) terms,
- \( n \) is the number of terms,
- \( a_1 \) is the first term,
- \( a_n \) is the \( n \)-th term.

We are given:
- \( a_{36} = 14 \) (the 36th term),
- \( d = \frac{1}{8} \) (the common difference),
- \( n = 36 \).

To use the formula, we need to find \( a_1 \) (the first term). The \( n \)-th term of an arithmetic sequence can be expressed in terms of the first term and the common difference:

\[
a_n = a_1 + (n-1)d
\]

Substituting \( n = 36 \):

\[
a_{36} = a_1 + (36-1)d = a_1 + 35 \cdot \frac{1}{8}
\]

Now we can set this equal to 14:

\[
14 = a_1 + \frac{35}{8}
\]

To solve for \( a_1 \), first, convert 14 to a fraction with the same denominator:

\[
14 = \frac{112}{8}
\]

Now we can substitute:

\[
\frac{112}{8} = a_1 + \frac{35}{8}
\]

Subtract \( \frac{35}{8} \) from both sides:

\[
a_1 = \frac{112}{8} - \frac{35}{8} = \frac{112 - 35}{8} = \frac{77}{8}
\]

Now we have \( a_1 \) and we can calculate \( S_{36} \):

Next, we need to calculate \( a_{36} \):

\[
S_{36} = \frac{36}{2} \times \left(a_1 + a_{36}\right) = 18 \times \left(\frac{77}{8} + 14\right)
\]

Again, convert 14 to a fraction with a common denominator:

\[
14 = \frac{112}{8}
\]

Now combine:

\[
S_{36} = 18 \times \left(\frac{77}{8} + \frac{112}{8}\right) = 18 \times \frac{189}{8}
\]

Calculate \( 18 \times 189 = 3402 \):

\[
S_{36} = \frac{3402}{8}
\]

Now simplify \( \frac{3402}{8} \):

\[
3402 \div 8 = 425.25
\]

Thus, the sum of the first 36 terms of the arithmetic sequence is \( \boxed{425.25} \).