Math Problem Statement

The sums of the 5th and 13th terms of an Arithmetic Progression are 21 and 37 respectively. Determine: i) The 40th term. (3 Marks) ii) The sum of the 29th term. (3 Marks)

Solution

Let's solve the problem step by step.

Given:

  • The sum of the 5th and 13th terms of an Arithmetic Progression (AP) is 21.
  • The sum of the 13th and 21st terms is 37.

Step 1: Express the terms in terms of the first term aa and the common difference dd.

  • The general formula for the nn-th term of an AP is given by: Tn=a+(n1)dT_n = a + (n-1)d

  • For the 5th term T5T_5: T5=a+4dT_5 = a + 4d

  • For the 13th term T13T_{13}: T13=a+12dT_{13} = a + 12d

  • For the 21st term T21T_{21}: T21=a+20dT_{21} = a + 20d

Step 2: Set up the equations based on the problem statement.

  • The sum of the 5th and 13th terms: T5+T13=(a+4d)+(a+12d)=2a+16d=21(Equation 1)T_5 + T_{13} = (a + 4d) + (a + 12d) = 2a + 16d = 21 \quad \text{(Equation 1)}

  • The sum of the 13th and 21st terms: T13+T21=(a+12d)+(a+20d)=2a+32d=37(Equation 2)T_{13} + T_{21} = (a + 12d) + (a + 20d) = 2a + 32d = 37 \quad \text{(Equation 2)}

Step 3: Solve the system of equations.

  1. From Equation 1: 2a+16d=21(Equation 1)2a + 16d = 21 \quad \text{(Equation 1)} Divide by 2: a+8d=10.5(Equation 3)a + 8d = 10.5 \quad \text{(Equation 3)}

  2. From Equation 2: 2a+32d=37(Equation 2)2a + 32d = 37 \quad \text{(Equation 2)} Divide by 2: a+16d=18.5(Equation 4)a + 16d = 18.5 \quad \text{(Equation 4)}

Step 4: Subtract Equation 3 from Equation 4.

(a+16d)(a+8d)=18.510.5(a + 16d) - (a + 8d) = 18.5 - 10.5 8d=88d = 8 d=1d = 1

Step 5: Substitute d=1d = 1 back into Equation 3 to find aa.

a+8(1)=10.5a + 8(1) = 10.5 a+8=10.5a + 8 = 10.5 a=2.5a = 2.5

Step 6: Determine the required values.

i) The 40th term T40T_{40}:

T40=a+39d=2.5+39(1)=2.5+39=41.5T_{40} = a + 39d = 2.5 + 39(1) = 2.5 + 39 = 41.5

ii) The sum of the first 29 terms S29S_{29}:

The sum of the first nn terms of an AP is given by: Sn=n2×(2a+(n1)d)S_n = \frac{n}{2} \times (2a + (n-1)d)

For n=29n = 29: S29=292×(2(2.5)+28(1))=292×(5+28)=292×33=9572=478.5S_{29} = \frac{29}{2} \times (2(2.5) + 28(1)) = \frac{29}{2} \times (5 + 28) = \frac{29}{2} \times 33 = \frac{957}{2} = 478.5

Final Answers:

i) The 40th term is 41.5.

ii) The sum of the first 29 terms is 478.5.

Would you like details or have any questions?


Here are 8 related questions you might find interesting:

  1. How do you find the common difference in an AP if only the first term and the 10th term are given?
  2. What is the formula for the sum of the first nn terms of an AP?
  3. How does changing the common difference affect the terms in an AP?
  4. Can the common difference in an AP be a fraction?
  5. What happens if the common difference is negative?
  6. How would you calculate the 50th term of an AP?
  7. What is the difference between an arithmetic progression and a geometric progression?
  8. How can you determine if a given sequence is an AP?

Tip: When solving arithmetic progression problems, always write down the general formula for the nn-th term and the sum of the first nn terms. This helps in setting up equations quickly.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Progression (AP)
Sum of AP terms
Nth term of AP
Sum of first n terms of AP

Formulas

Nth term of AP: T_n = a + (n-1)d
Sum of first n terms of AP: S_n = \frac{n}{2} \times (2a + (n-1)d)

Theorems

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Suitable Grade Level

High School