Math Problem Statement

Consider the arithmetic sequence 12, 19, 26, 33, .... a. Find the 8th term of the sequence. b. Find the sum of the first 10 terms of the sequence. c. The sum of the first n terms is 915. Find the value of n.

Solution

Let's solve the problems related to the given arithmetic sequence:

The arithmetic sequence provided is 12,19,26,33,12, 19, 26, 33, \ldots.

Part (a): Find the 8th term of the sequence.

To find the nnth term of an arithmetic sequence, we use the formula: an=a1+(n1)da_n = a_1 + (n-1) \cdot d where:

  • ana_n is the nnth term,
  • a1a_1 is the first term,
  • dd is the common difference.

Given:

  • a1=12a_1 = 12,
  • The common difference d=1912=7d = 19 - 12 = 7.

For the 8th term (n=8n = 8): a8=12+(81)7=12+77=12+49=61a_8 = 12 + (8-1) \cdot 7 = 12 + 7 \cdot 7 = 12 + 49 = 61

So, the 8th term is 61.

Part (b): Find the sum of the first 10 terms of the sequence.

The sum of the first nn terms of an arithmetic sequence is given by: Sn=n2(2a1+(n1)d)S_n = \frac{n}{2} \cdot (2a_1 + (n-1) \cdot d)

For the first 10 terms (n=10n = 10): S10=102(212+(101)7)S_{10} = \frac{10}{2} \cdot (2 \cdot 12 + (10-1) \cdot 7) S10=5(24+63)=587=435S_{10} = 5 \cdot (24 + 63) = 5 \cdot 87 = 435

So, the sum of the first 10 terms is 435.

Part (c): The sum of the first nn terms is 915. Find the value of nn.

We use the sum formula again, but now solve for nn: Sn=n2(2a1+(n1)d)S_n = \frac{n}{2} \cdot (2a_1 + (n-1) \cdot d) Given: 915=n2(24+(n1)7)915 = \frac{n}{2} \cdot (24 + (n-1) \cdot 7) 915=n2(24+7n7)915 = \frac{n}{2} \cdot (24 + 7n - 7) 915=n2(7n+17)915 = \frac{n}{2} \cdot (7n + 17) Multiply both sides by 2 to eliminate the fraction: 1830=n(7n+17)1830 = n \cdot (7n + 17) Expanding and rearranging the equation: 7n2+17n1830=07n^2 + 17n - 1830 = 0 This is a quadratic equation. To solve it, we can use the quadratic formula: n=b±b24ac2an = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=7a = 7, b=17b = 17, and c=1830c = -1830.

n=17±172+47183027n = \frac{-17 \pm \sqrt{17^2 + 4 \cdot 7 \cdot 1830}}{2 \cdot 7} n=17±289+5124014n = \frac{-17 \pm \sqrt{289 + 51240}}{14} n=17±5152914n = \frac{-17 \pm \sqrt{51529}}{14} n=17±22714n = \frac{-17 \pm 227}{14} Taking the positive root: n=21014=15n = \frac{210}{14} = 15

So, n=15n = 15.

If you have any questions or need further details, feel free to ask!

Here are 5 related questions:

  1. How would the common difference change the sequence if it were 5 instead of 7?
  2. What is the 20th term of this arithmetic sequence?
  3. How do you find the sum of the first 50 terms of an arithmetic sequence?
  4. Can you derive the formula for the sum of an arithmetic sequence from first principles?
  5. How does the sum formula change if the sequence starts at a different term?

Tip: When dealing with sequences, always check the pattern of the common difference (arithmetic) or ratio (geometric) to ensure you're applying the correct formula.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Arithmetic Sequences
Algebra
Quadratic Equations

Formulas

n-th term of an arithmetic sequence: a_n = a_1 + (n-1) * d
Sum of the first n terms of an arithmetic sequence: S_n = (n/2) * (2a_1 + (n-1) * d)
Quadratic equation formula: ax^2 + bx + c = 0

Theorems

Quadratic Formula

Suitable Grade Level

Grades 8-10