Math Problem Statement

Solution

To solve this problem, we need to analyze the given arithmetic sequence a1,a2,,a2m+1a_1, a_2, \ldots, a_{2m+1} and use the provided properties. Let's break down the information:

Given properties:

  1. a1+a3++a2m+1=132a_1 + a_3 + \cdots + a_{2m+1} = 132
  2. a2+a4++a2m=120a_2 + a_4 + \cdots + a_{2m} = 120

Since this is an arithmetic sequence, each term can be expressed as ai=a+(i1)da_i = a + (i-1)d, where aa is the first term and dd is the common difference.

Step-by-Step Solution:

  1. Sum of Odd-Indexed Terms: The terms involved are a1,a3,a5,,a2m+1a_1, a_3, a_5, \ldots, a_{2m+1}.

    • These are m+1m+1 terms.
    • Each term can be written as a2k1=a+(2k2)da_{2k-1} = a + (2k-2)d for k=1k = 1 to m+1m+1.

    The sum is: k=1m+1[a+(2k2)d]=(m+1)a+dk=1m+1(2k2)\sum_{k=1}^{m+1} \left[a + (2k-2)d\right] = (m+1)a + d \sum_{k=1}^{m+1} (2k-2) The sum of the series 2k22k-2 from k=1k=1 to m+1m+1 is: 2k=1m+1(k1)=2[(m+1)m2]=m(m+1)2 \sum_{k=1}^{m+1} (k-1) = 2 \left[\frac{(m+1)m}{2}\right] = m(m+1) Therefore: (m+1)a+dm(m+1)=132(m+1)a + d \cdot m(m+1) = 132

  2. Sum of Even-Indexed Terms: The terms involved are a2,a4,a6,,a2ma_2, a_4, a_6, \ldots, a_{2m}.

    • These are mm terms.
    • Each term can be written as a2k=a+(2k1)da_{2k} = a + (2k-1)d for k=1k = 1 to mm.

    The sum is: k=1m[a+(2k1)d]=ma+dk=1m(2k1)\sum_{k=1}^{m} \left[a + (2k-1)d\right] = ma + d \sum_{k=1}^{m} (2k-1) The sum of the series 2k12k-1 from k=1k=1 to mm is: k=1m(2k1)=m2\sum_{k=1}^{m} (2k-1) = m^2 Therefore: ma+dm2=120ma + d \cdot m^2 = 120

  3. Solving the System of Equations: We have two equations: (m+1)a+dm(m+1)=132(1)(m+1)a + d \cdot m(m+1) = 132 \quad \text{(1)} ma+dm2=120(2)ma + d \cdot m^2 = 120 \quad \text{(2)}

    Divide equation (1) by m+1m+1: a+dm=132m+1a + d \cdot m = \frac{132}{m+1}

    Divide equation (2) by mm: a+dm=120ma + d \cdot m = \frac{120}{m}

    Since the left-hand sides are equal: 132m+1=120m\frac{132}{m+1} = \frac{120}{m} Cross-multiplying gives: 132m=120(m+1)132m = 120(m+1) 132m=120m+120132m = 120m + 120 12m=12012m = 120 m=10m = 10

Thus, the value of mm is 1010.

Would you like further details or have any questions?

Here are 5 related questions you might be interested in:

  1. How do you find the common difference in an arithmetic sequence?
  2. What is the general formula for the sum of an arithmetic series?
  3. How do you solve a system of linear equations?
  4. Can you explain how to derive the formula for the sum of odd or even indexed terms in a sequence?
  5. What are some applications of arithmetic sequences in real life?

Tip: Always verify your solutions by substituting them back into the original equations to ensure they satisfy all given conditions.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Sequences
Series Summation

Formulas

Arithmetic Sequence Formula: a_n = a_1 + (n-1)d
Sum of Arithmetic Series: S_n = n/2 * (a_1 + a_n)

Theorems

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Suitable Grade Level

High School