The sum of the first six terms of an arithmetic progression (AP) is 42. The ratio of its 10th term to its 30th term is 1:3. Calculate the first and the thirteenth term of the AP.

Question

19-08-2024
Mathematics

High School

We appreciate your visit to The sum of the first six terms of an arithmetic progression AP is 42 The ratio of its 10th term to its 30th term is. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

The sum of the first six terms of an arithmetic progression (AP) is 42. The ratio of its 10th term to its 30th term is 1:3. Calculate the first and the thirteenth term of the AP.

Answer :

Other Questions

Is Beowulf a humble or arrogant type of heroic figure?

The stopping distance (at some fixed speed) of regular tires on glare

### Instructions: Read each question carefully. For each item, choose the correct

What time is 5 3/4 hours after 11:32 PM?

Gas Laws Fact Sheet [tex] \[ \begin{array}{|l|l|} \hline \text{Ideal gas law} &

The quotient of [tex]\left(x^4+5x^3-3x-15\right)[/tex] and [tex]\left(x^3-3\right)[/tex] is a polynomial. What is the

### Exercise 1: Put the verbs in brackets into the present perfect

You pull with a force of 77 N on a piece of

Barbara is converting [tex]78^{\circ} F[/tex] to degrees Celsius. First, she subtracts 32

At the beginning of a race, someone asks a group of runners

KyleEvans KyleEvans · Answer 1 · 2025-02-10T05:24:50-05:00

Final answer:

To find the first and thirteenth term of an arithmetic progression (AP), we can use the sum of the first six terms and the ratio of the 10th term to the 30th term. By solving a system of equations, we find that the first term is 7 and the thirteenth term is 31.

Explanation:

To find the first term of the arithmetic progression (AP), we need to use the formula for the sum of the first n terms. In this case, the sum of the first six terms is 42, so we can write the equation 42 = (n/2)(2a + (n-1)d), where n is the number of terms, a is the first term, and d is the common difference. Substituting n=6 and rearranging the equation, we get 6a + 15d = 84.

Next, we can use the ratio of the 10th term to the 30th term to find the common difference. The ratio is given as 1:3, so we can write the equation a + 9d = (1/3)(a + 29d). Rearranging the equation gives us 8a - 60d = 0.

Solving the system of equations 6a + 15d = 84 and 8a - 60d = 0, we find that a = 7 and d = 2. Therefore, the first term of the AP is 7 and the thirteenth term can be found using the formula tn = a + (n-1)d. Plugging in n=13, a=7, and d=2, we get t13 = 7 + 12(2) = 31.