Math Problem Statement

solve for arithmetic progression

Find the 7th term of an A.P. if the 3rd term is 10 and the 10th term is -4. Find the 15th term of an A.P.; the 31st term is 40 and the sum of the 3rd and 11th terms is 16.

Solution

Let's work through the two problems related to Arithmetic Progressions (A.P.) shown in the image.


Problem 1:

Find the 7th term of an A.P. if the 3rd term is 10, and the 10th term is -4.

Solution:

The general formula for the nn-th term of an arithmetic progression is: an=a+(n1)da_n = a + (n-1)d where:

  • aa is the first term,
  • dd is the common difference,
  • nn is the term number.

From the problem:

  • The 3rd term (a3a_3) is 10, so: a+2d=10(1)a + 2d = 10 \quad \text{(1)}
  • The 10th term (a10a_{10}) is -4, so: a+9d=4(2)a + 9d = -4 \quad \text{(2)}

Now, subtract equation (1) from equation (2): (a+9d)(a+2d)=410(a + 9d) - (a + 2d) = -4 - 10 7d=147d = -14 d=2d = -2

Now substitute d=2d = -2 into equation (1): a+2(2)=10a + 2(-2) = 10 a4=10a - 4 = 10 a=14a = 14

Now, find the 7th term (a7a_7): a7=a+6d=14+6(2)=1412=2a_7 = a + 6d = 14 + 6(-2) = 14 - 12 = 2

Thus, the 7th term is 2.


Problem 2:

Find the 15th term of an A.P.; the 31st term is 40, and the sum of the 3rd and 11th terms is 16.

Solution:

From the general formula for the nn-th term: an=a+(n1)da_n = a + (n-1)d We are given:

  • The 31st term (a31a_{31}) is 40: a+30d=40(1)a + 30d = 40 \quad \text{(1)}
  • The sum of the 3rd term (a3a_3) and the 11th term (a11a_{11}) is 16: (a+2d)+(a+10d)=16(a + 2d) + (a + 10d) = 16 2a+12d=16(2)2a + 12d = 16 \quad \text{(2)}

From equation (2), we can simplify: a+6d=8(3)a + 6d = 8 \quad \text{(3)}

Now, subtract equation (3) from equation (1): (a+30d)(a+6d)=408(a + 30d) - (a + 6d) = 40 - 8 24d=3224d = 32 d=3224=43d = \frac{32}{24} = \frac{4}{3}

Substitute d=43d = \frac{4}{3} into equation (3): a+6(43)=8a + 6\left(\frac{4}{3}\right) = 8 a+8=8a + 8 = 8 a=0a = 0

Now, find the 15th term (a15a_{15}): a15=a+14d=0+14(43)=56318.67a_{15} = a + 14d = 0 + 14\left(\frac{4}{3}\right) = \frac{56}{3} \approx 18.67

Thus, the 15th term is approximately 18.67.


Do you have any questions or would you like more details on these solutions? Here are five related questions for practice:

  1. What is the general formula for the sum of the first nn terms of an arithmetic progression?
  2. If the common difference of an A.P. is positive, how does it affect the progression?
  3. Can the common difference of an arithmetic progression be zero?
  4. How would you find the 50th term of an A.P. if the 5th term is 12 and the 10th term is 22?
  5. What are the necessary steps to find the common difference of an arithmetic progression given any two terms?

Tip: Always write down what is given and the general formula first; it helps to structure your approach in arithmetic progressions.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Progression
Algebra

Formulas

nth term of an A.P.: a_n = a + (n-1)d
Sum of terms: S_n = (n/2) * (2a + (n-1)d)

Theorems

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Suitable Grade Level

Grades 9-12