The 7th term of an AP is twice the third term;and the sum of the first four term,find:
[1]common ratio
[2]find the sum of the first 12th
terms
First of all, in an AP we have a common difference,
the common ratio is found in a geometric sequence (GP).
Besides that, you have missing information.
Fix your post
To solve this problem, let's assume that the first term of the arithmetic progression (AP) is denoted by "a", and the common difference between the terms is denoted by "d".
Given:
The 7th term = 2 times the 3rd term
Step 1: Find the common difference (d):
Using the given information, we can write the 7th term as:
a + 6d
And the 3rd term as:
a + 2d
According to the problem, the 7th term is twice the 3rd term:
a + 6d = 2(a + 2d)
Expand and simplify the equation:
a + 6d = 2a + 4d
6d - 4d = 2a - a
2d = a
So, we have found the relation between the first term (a) and the common difference (d).
[1] The common ratio is 2d.
Step 2: Find the sum of the first four terms:
The sum of the first four terms of an AP can be calculated using the formula:
Sum = (n/2)(2a + (n - 1)d)
In this case, n = 4, a = a, and d = d.
The formula becomes:
Sum = (4/2)(2a + (4 - 1)d)
Sum = 2(2a + 3d)
Sum = 4a + 6d
[2] Find the sum of the first 12 terms:
The sum of the first twelve terms can be calculated using the same formula, but with n = 12:
Sum = (12/2)(2a + (12 - 1)d)
Sum = 6(2a + 11d)
Sum = 12a + 66d
So, the sum of the first 12 terms is 12a + 66d.
To find the answers to the given question, we'll have to use the formula for the nth term of an arithmetic progression (AP). The formula is:
nth term = a + (n - 1)d
where "a" represents the first term, "n" represents the position of the term, and "d" represents the common difference between each term.
Let's break down the given information for this specific AP:
1. The 7th term of the AP is twice the third term. This can be represented as:
7th term = 2 * 3rd term
Using the formula, we can write the two equations as:
a + (7 - 1)d = 2(a + (3 - 1)d)
a + 6d = 2(a + 2d)
2. The sum of the first four terms. This can be represented as:
Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d)
Now, let's solve the equations step by step:
1. To find the common difference (d), we'll simplify the equation:
a + 6d = 2a + 4d
Rearranging the terms, we get:
6d - 4d = 2a - a
Simplifying further, we have:
2d = a
So, the common ratio (d) is equal to half of the first term (a).
2. To find the sum of the first 12 terms, we'll use the formula for the sum of an AP:
Sum of n terms = (n/2)(2a + (n - 1)d)
Plugging in the values, we get:
Sum of 12 terms = (12/2)(2a + (12 - 1)d)
= 6(2a + 11d)
= 12a + 66d
Since we know that d = 2a, we can substitute the value and simplify:
Sum of 12 terms = 12a + 66(2a)
= 12a + 132a
= 144a
Therefore, the sum of the first 12 terms is 144 times the first term (a).