The 9th and the 22nd term is 29 and 55 respectively. Find the sum of the first 60 terms
doesn't specify arithmetic (or geometric) , but it comes out even
d = (55 - 29) / (22 - 9)
1st term = 29 - (9 d)
60th term = 1st term + (59 d)
sum = (1st term + 60th term) * (60 / 2)
you know that
d = (55-29)/(22-9) = 26/13 = 2
a = a9 - 8d = 29-16 = 13
S60 = 60/2 (2*13 + 59*2) = _____
oops ... 1st term = 29 - (8 d) ... not (9 d)
To find the sum of the first 60 terms, we need to find the common difference in the arithmetic sequence, and then use the formula for the sum of an arithmetic sequence.
First, we need to find the common difference (d) in the sequence. The common difference is the constant difference between the terms in an arithmetic sequence.
The 9th term is 29, and the 22nd term is 55.
We can find the common difference by subtracting the 9th term from the 22nd term and dividing by the number of terms between them:
d = (55 - 29) / (22 - 9)
d = 26 / 13
d = 2
Now that we know the common difference is 2, we can use the formula for the sum of an arithmetic sequence:
Sum = (n/2)(2a + (n-1)d)
In this formula:
- n is the number of terms (in this case, 60)
- a is the first term (we need to find this)
- d is the common difference (2)
To find the first term (a), we can use either the formula for the nth term in an arithmetic sequence or use the given information.
Let's use the given information:
The 9th term is 29, so we can substitute n = 9 and a = 29 into the formula for the nth term to solve for a:
a + (9-1)d = 29
a + 8d = 29
a + 8(2) = 29
a + 16 = 29
a = 29 - 16
a = 13
Now that we have the first term (a = 13) and the common difference (d = 2), we can substitute these values into the formula for the sum of an arithmetic sequence:
Sum = (60/2)(2a + (60-1)d)
Sum = 30(2(13) + 59(2))
Sum = 30(26 + 118)
Sum = 30(144)
Sum = 4320
Therefore, the sum of the first 60 terms is 4320.