Math Problem Statement

The number of terms in Arithmetic sequence is even . The sum of odd terms is 24.if the sum of even terms is 30 and the last term exceeds the first 10.5 ,then find the number of terms ,the first term ,the common difference

Solution

Let's break down the information given:

  1. The number of terms in the arithmetic sequence is even.
  2. The sum of the odd terms is 24.
  3. The sum of the even terms is 30.
  4. The last term exceeds the first term by 10.5.

Let the number of terms in the sequence be 2n2n. This is because the number of terms is even, so we denote the number of odd terms and even terms separately.

  • Let the first term of the arithmetic sequence be aa.
  • Let the common difference be dd.

The odd terms form their own arithmetic sequence:

  • First odd term: aa,
  • Second odd term: a+2da + 2d,
  • Third odd term: a+4da + 4d, and so on.

There are nn odd terms, and their sum is given as 24. So, the sum of these nn odd terms is: Sodd=n(2a+(n1)2d2)=24S_{\text{odd}} = n \cdot \left( \frac{2a + (n-1) \cdot 2d}{2} \right) = 24 Simplifying this: Sodd=n(a+(n1)d)=24(Equation 1)S_{\text{odd}} = n \left( a + (n-1) \cdot d \right) = 24 \quad \text{(Equation 1)}

The even terms also form an arithmetic sequence:

  • First even term: a+da + d,
  • Second even term: a+3da + 3d,
  • Third even term: a+5da + 5d, and so on.

There are nn even terms, and their sum is given as 30. So, the sum of these nn even terms is: Seven=n(2(a+d)+(n1)2d2)=30S_{\text{even}} = n \cdot \left( \frac{2(a + d) + (n-1) \cdot 2d}{2} \right) = 30 Simplifying this: Seven=n((a+d)+(n1)d)=30(Equation 2)S_{\text{even}} = n \left( (a + d) + (n-1) \cdot d \right) = 30 \quad \text{(Equation 2)}

Finally, we know that the last term exceeds the first term by 10.5. The last term of the sequence (which is the 2n2n-th term) is given by: a+(2n1)d=a+10.5(Equation 3)a + (2n - 1) \cdot d = a + 10.5 \quad \text{(Equation 3)}

We now have three equations:

  1. n(a+(n1)d)=24n \left( a + (n-1) \cdot d \right) = 24
  2. n((a+d)+(n1)d)=30n \left( (a + d) + (n-1) \cdot d \right) = 30
  3. a+(2n1)d=a+10.5a + (2n - 1) \cdot d = a + 10.5

Let's solve these equations step by step.The solution to the system of equations is:

  • The number of terms 2n2n is 2×4=82 \times 4 = 8.
  • The first term aa is 1.51.5.
  • The common difference dd is 1.51.5.

Thus, the arithmetic sequence has 8 terms, with the first term being 1.51.5 and the common difference being 1.51.5.

Would you like a detailed verification or further explanation on how these values fit the given conditions? Here are 5 related questions:

  1. How is the sum of the terms in an arithmetic sequence calculated?
  2. Can you find the explicit form of the sequence using the values of aa and dd?
  3. What happens to the sequence if the common difference is negative?
  4. How would you solve this problem if only the sum of all terms were given?
  5. Can you explore sequences where the common difference is a fraction?

Tip: In any arithmetic sequence, knowing the first term and the common difference allows you to easily construct any term in the sequence!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Arithmetic Sequences
Summation of Series
System of Equations

Formulas

Sum of an arithmetic sequence: S_n = n/2 * (first_term + last_term)
nth term of an arithmetic sequence: a_n = a + (n-1)d
Sum of odd terms: S_odd = n * (a + (n-1)d)
Sum of even terms: S_even = n * ((a + d) + (n-1)d)

Theorems

Properties of Arithmetic Sequences

Suitable Grade Level

Grades 9-12