Math Problem Statement

Which of the following is true?

A.	The tenth term of the sequence 4, 7, 10, 13, … is 30.
B.	an = 11 + 6n and an={11,if n=1an−1+6n,if n>1

both define the same sequence. C. The sum of the first 12 terms of the arithmetic series with a1 = −5 and a12 = 232 is 39. D. ∑20i=1(3i−15)=450

Solution

Let's analyze each statement one by one:

A. The tenth term of the sequence 4, 7, 10, 13, … is 30.

This is an arithmetic sequence where the first term a1=4a_1 = 4 and the common difference d=3d = 3. The formula for the nth term of an arithmetic sequence is:

an=a1+(n1)da_n = a_1 + (n - 1) d

For n=10n = 10, we have:

a10=4+(101)3=4+93=4+27=31a_{10} = 4 + (10 - 1) \cdot 3 = 4 + 9 \cdot 3 = 4 + 27 = 31

So, the tenth term is 31, not 30. A is false.


**B. an=11+6na_n = 11 + 6n and (a_n = \begin{cases}

11 & \text{if } n = 1 \ a_{n-1} + 6n & \text{if } n > 1 \end{cases}) both define the same sequence.**

Let's compare both definitions.

  1. For an=11+6na_n = 11 + 6n:

    For n=1n = 1: a1=11+6×1=17a_1 = 11 + 6 \times 1 = 17
    For n=2n = 2: a2=11+6×2=23a_2 = 11 + 6 \times 2 = 23
    For n=3n = 3: a3=11+6×3=29a_3 = 11 + 6 \times 3 = 29

  2. For the recursive sequence (a_n = \begin{cases} 11 & \text{if } n = 1 \ a_{n-1} + 6n & \text{if } n > 1 \end{cases}):

    For n=1n = 1: a1=11a_1 = 11
    For n=2n = 2: a2=a1+6×2=11+12=23a_2 = a_1 + 6 \times 2 = 11 + 12 = 23
    For n=3n = 3: a3=a2+6×3=23+18=41a_3 = a_2 + 6 \times 3 = 23 + 18 = 41

Clearly, the sequences are different. So, B is false.


C. The sum of the first 12 terms of the arithmetic series with a1=5a_1 = −5 and a12=232a_{12} = 232 is 39.

The formula for the sum of the first nn terms of an arithmetic series is:

Sn=n2(a1+an)S_n = \frac{n}{2} (a_1 + a_n)

For n=12n = 12, a1=5a_1 = -5, and a12=232a_{12} = 232:

S12=122(5+232)=6×227=1362S_{12} = \frac{12}{2} (-5 + 232) = 6 \times 227 = 1362

The sum is 1362, not 39. So, C is false.


D. i=120(3i15)=450\sum_{i=1}^{20} (3i - 15) = 450

The sum can be rewritten as:

i=120(3i15)=i=1203ii=12015\sum_{i=1}^{20} (3i - 15) = \sum_{i=1}^{20} 3i - \sum_{i=1}^{20} 15

The sum of the first 20 terms of 3i3i is:

i=1203i=3i=120i=3×20(21)2=3×210=630\sum_{i=1}^{20} 3i = 3 \sum_{i=1}^{20} i = 3 \times \frac{20(21)}{2} = 3 \times 210 = 630

The sum of 1515 repeated 20 times is:

i=12015=15×20=300\sum_{i=1}^{20} 15 = 15 \times 20 = 300

Now, subtract:

630300=330630 - 300 = 330

So, the sum is 330, not 450. D is false.


Conclusion:

All four statements are false.


Do you want further details or have any questions?

Here are 5 related questions to explore:

  1. How do you find the sum of an arithmetic sequence?
  2. What is the difference between an explicit and recursive formula for sequences?
  3. How do you calculate the nth term of an arithmetic sequence?
  4. How can you decompose summations like i=1n(ai+b)\sum_{i=1}^{n} (ai + b)?
  5. How do you identify if two sequences are equivalent?

Tip: When working with summations, break them into simpler parts to handle constants and linear terms separately.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Sequences
Recursive Sequences
Summation
Arithmetic Series

Formulas

nth term of an arithmetic sequence: an = a1 + (n - 1)d
Sum of an arithmetic series: Sn = n/2(a1 + an)
Summation of linear terms: ∑i=1n(ai + b) = a∑i=1ni + b·n

Theorems

Arithmetic sequence formula
Summation of an arithmetic series

Suitable Grade Level

Grades 10-12