Math Problem Statement

The first term of an arithmetic sequence is equal to 3. Find the value of the 50th term. Find the 50th term and the sum of all 3-digit odd positive integers.

Solution

Let's break down the problems from the image:

Problem 1:

The first term of an arithmetic sequence is equal to 3. Find the value of the 50th term.

To find the 50th term of an arithmetic sequence, we use the formula for the nth term of an arithmetic sequence: an=a1+(n1)da_n = a_1 + (n - 1) \cdot d Where:

  • ana_n is the nth term
  • a1a_1 is the first term (3 in this case)
  • dd is the common difference (which is not provided)
  • nn is the position of the term (50 in this case)

Without the common difference dd, we can't solve this fully yet. If you provide the common difference, I can calculate the 50th term.


Problem 2:

Find the 50th term and the sum of all 3-digit odd positive integers.

First, we need to find the 50th term of the sequence of 3-digit odd positive integers.

  • The first term of this sequence is 101.
  • The common difference dd is 2 (since each odd integer increases by 2).

To find the 50th term: a50=101+(501)2=101+492=101+98=199a_{50} = 101 + (50 - 1) \cdot 2 = 101 + 49 \cdot 2 = 101 + 98 = 199 So, the 50th term is 199.

Next, we calculate the sum of all 3-digit odd integers.

  • The first term (a1a_1) is 101.
  • The last term (ana_n) is 999.
  • The common difference d=2d = 2.

The formula for the sum of an arithmetic sequence is: Sn=n2(a1+an)S_n = \frac{n}{2} \cdot (a_1 + a_n) We need to first find how many terms (nn) are in the sequence. To do this, we solve for nn using the nth-term formula: an=a1+(n1)da_n = a_1 + (n-1) \cdot d Substitute an=999a_n = 999, a1=101a_1 = 101, and d=2d = 2: 999=101+(n1)2999 = 101 + (n - 1) \cdot 2 999101=(n1)2999 - 101 = (n - 1) \cdot 2 898=(n1)2898 = (n - 1) \cdot 2 n1=8982=449n - 1 = \frac{898}{2} = 449 n=450n = 450 So, there are 450 terms in this sequence.

Now, apply the sum formula: S450=4502(101+999)=2251100=247500S_{450} = \frac{450}{2} \cdot (101 + 999) = 225 \cdot 1100 = 247500

Summary:

  • 50th term of the 3-digit odd sequence = 199
  • Sum of all 3-digit odd positive integers = 247,500

Let me know if you want further details or clarifications! Here are some related questions:

  1. What is the general form of an arithmetic sequence?
  2. How would you find the sum of an arithmetic sequence if only the common difference is known?
  3. What would happen to the sum if the sequence had been geometric instead of arithmetic?
  4. How would you solve for the sum of an arithmetic series if you only knew the first few terms?
  5. How would you calculate the sum of all 4-digit even integers?

Tip: Always verify if all necessary parameters (like the common difference) are provided when solving sequence problems.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Sequences
Series

Formulas

nth term of an arithmetic sequence: a_n = a_1 + (n - 1) * d
Sum of an arithmetic sequence: S_n = (n / 2) * (a_1 + a_n)

Theorems

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Suitable Grade Level

Grades 7-9