An arithmetic progression has 11 as its first term and fourth term is 32.find the sum of the first nine terms.
351
3 differences between 1st and 4th terms ... d = (32 - 11) / 3 = 7
9th term = 1st term + 8 d = 67
sum of 1st 9 terms ... (11 + 67) * (9/2)
To find the sum of the first nine terms of an arithmetic progression, we need to find the common difference (d) and then use the formula for the sum of an arithmetic progression.
Step 1: Find the common difference (d)
The fourth term is given as 32, and the first term is 11.
So, the difference between these two terms will give us the common difference (d).
d = fourth term - first term = 32 - 11 = 21
Step 2: Find the sum of the first nine terms
The formula for the sum of an arithmetic progression is given by:
Sn = (n/2)(2a + (n-1)d)
Where Sn is the sum of the first n terms, a is the first term, n is the number of terms, and d is the common difference.
Substituting the values into the formula:
n = 9 (as we need to find the sum of the first nine terms)
a = 11 (first term)
d = 21 (common difference)
Sn = (9/2)(2 * 11 + (9-1) * 21)
= (9/2)(22 + 8 * 21)
= (9/2)(22 + 168)
= (9/2)(190)
= 9 * 95
= 855
Therefore, the sum of the first nine terms of the arithmetic progression is 855.
To find the sum of the first nine terms of an arithmetic progression, you can use the formula for the sum of an arithmetic series:
Sn = (n/2) * (2a + (n-1)d)
Where:
- Sn is the sum of the first n terms
- a is the first term
- d is the common difference between consecutive terms
- n is the number of terms
In this case, we know the first term (a = 11), and the fourth term (a3 = 32). We need to find the common difference (d) to proceed.
To find the common difference (d), we can use the formula for the nth term of an arithmetic progression:
an = a + (n-1)d
Substituting the known values, we have:
32 = 11 + (4-1)d
Simplifying the equation, we get:
32 = 11 + 3d
Subtracting 11 from both sides, we have:
3d = 21
Dividing both sides by 3, we find:
d = 7
Now that we have the common difference (d = 7), we can proceed to find the sum of the first nine terms (Sn).
Using the formula for the sum of an arithmetic series, we substitute the values:
Sn = (n/2) * (2a + (n-1)d)
= (9/2) * (2*11 + (9-1) * 7)
= (9/2) * (22 + 8*7)
= (9/2) * (22 + 56)
= (9/2) * 78
= 351
Therefore, the sum of the first nine terms of the arithmetic progression is 351.