Math Problem Statement

The first five terms of a sequence are 1, -4, -9, -14, -19. (a) Write down an expression for the nth term of the sequence. (b) Is -110 a term of this sequence? (c) Find the value of n when the sum of the first n terms is -318.

Solution

Let's break down and solve each part of the question as shown in the image.

(a) Write down an expression for the nth term of the sequence:

The sequence given is: 1,4,9,14,191, -4, -9, -14, -19. This appears to be an arithmetic sequence where each term is decreasing by a constant value. The common difference dd is: d=41=5d = -4 - 1 = -5 The first term a1=1a_1 = 1. The general formula for the nth term of an arithmetic sequence is: an=a1+(n1)da_n = a_1 + (n-1) \cdot d Substitute a1=1a_1 = 1 and d=5d = -5: an=1+(n1)(5)a_n = 1 + (n-1)(-5) Simplify: an=15n+5=5n+6a_n = 1 - 5n + 5 = -5n + 6 So, the nth term is: an=5n+6a_n = -5n + 6

(b) Is 110-110 a term of this sequence?

We need to determine if 110-110 can be a value of the nth term, i.e., we solve for nn in: 5n+6=110-5n + 6 = -110 First, subtract 6 from both sides: 5n=1106-5n = -110 - 6 5n=116-5n = -116 Now, divide by 5-5: n=1165=23.2n = \frac{-116}{-5} = 23.2 Since nn is not an integer, 110-110 is not a term of the sequence. Therefore, the answer is No.

(c) Find the value of nn when the sum of the first nn terms is 318-318:

The formula for the sum of the first nn terms of an arithmetic sequence is: Sn=n2[2a1+(n1)d]S_n = \frac{n}{2} [2a_1 + (n-1)d] Substitute Sn=318S_n = -318, a1=1a_1 = 1, and d=5d = -5: 318=n2[2(1)+(n1)(5)]-318 = \frac{n}{2} [2(1) + (n-1)(-5)] Simplify: 318=n2[25n+5]-318 = \frac{n}{2} [2 - 5n + 5] 318=n2[75n]-318 = \frac{n}{2} [7 - 5n] Multiply both sides by 2: 636=n(75n)-636 = n(7 - 5n) Expand the right-hand side: 636=7n5n2-636 = 7n - 5n^2 Rearrange to form a quadratic equation: 5n27n636=05n^2 - 7n - 636 = 0 Now, solve this quadratic equation using the quadratic formula: n=(7)±(7)24(5)(636)2(5)n = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(5)(-636)}}{2(5)} n=7±49+1272010n = \frac{7 \pm \sqrt{49 + 12720}}{10} n=7±1276910n = \frac{7 \pm \sqrt{12769}}{10} n=7±11310n = \frac{7 \pm 113}{10} So the two possible solutions are: n=7+11310=12orn=711310=10.6n = \frac{7 + 113}{10} = 12 \quad \text{or} \quad n = \frac{7 - 113}{10} = -10.6 Since nn must be a positive integer, n=12n = 12.

Summary of answers:

  • (a) The nth term of the sequence is an=5n+6a_n = -5n + 6.
  • (b) 110-110 is not a term of the sequence.
  • (c) The value of nn when the sum is 318-318 is n=12n = 12.

Do you have any questions or want further details on any part?

Here are some related questions for practice:

  1. What would the 15th term of this sequence be?
  2. How would the sequence change if the first term was 2 instead of 1?
  3. How do you derive the formula for the sum of an arithmetic sequence?
  4. What if you were asked for the sum of the first 20 terms instead?
  5. Can you determine a general rule for checking if a number is part of an arithmetic sequence?

Tip: Always check for the common difference first when identifying a sequence type!

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Math Problem Analysis

Mathematical Concepts

Arithmetic Sequence
Linear Equations
Summation of Arithmetic Sequences
Quadratic Equations

Formulas

nth term of arithmetic sequence: a_n = a_1 + (n-1) * d
Sum of arithmetic sequence: S_n = (n/2) * (2a_1 + (n-1) * d)
Quadratic formula: n = (-b ± sqrt(b^2 - 4ac)) / 2a

Theorems

Quadratic Formula Theorem

Suitable Grade Level

Grades 9-10