Find the sum of the first 12th terms of the sequence 2,5,8,11,
Full method pls
To find the sum of the first 12 terms of the sequence 2, 5, 8, 11, we can use the formula for the sum of an arithmetic series.
The formula for the sum of an arithmetic series is given by:
S = (n/2) * (a + l)
Where:
S = sum of the series
n = number of terms in the series
a = first term of the series
l = last term of the series
In this case, the first term (a) is 2 and the last term (l) can be found using the formula for the nth term of an arithmetic sequence:
l = a + (n - 1) * d
Where:
d = common difference between consecutive terms of the sequence
Since the common difference between consecutive terms is 3 (5 - 2 = 3), we can substitute these values into the formula:
l = 2 + (12 - 1) * 3
l = 2 + 33
l = 35
Now, we have the values for n, a, and l. We can plug these values into the formula for the sum of an arithmetic series to find the answer:
S = (12/2) * (2 + 35)
S = 6 * 37
S = 222
Therefore, the sum of the first 12 terms of the sequence 2, 5, 8, 11 is 222.
To find the sum of the first 12 terms of the sequence 2, 5, 8, 11, we need to find the formula for the nth term and then use it to calculate the sum.
From the given sequence, we can observe that the common difference between consecutive terms is 3. To find the formula for the nth term (an) of an arithmetic sequence, we use the formula:
an = a1 + (n - 1)d
Where:
an is the nth term
a1 is the first term
n is the number of terms
d is the common difference
Here, a1 = 2 and d = 3. Plugging these values into the formula, we get:
an = 2 + (n - 1)3
= 2 + 3n - 3
= 3n - 1
Now, we need to find the sum of the first 12 terms, S12, which can be calculated using the formula for the sum of an arithmetic series:
S12 = (n/2)(a1 + an)
Plugging in the values, we get:
S12 = (12/2)(2 + (3*12 - 1))
= 6(2 + 35)
= 6(37)
= 222
Therefore, the sum of the first 12 terms of the sequence 2, 5, 8, 11 is 222.
the common difference is 3
the 12th term is ... 2 + (11 * 3) = 35
sum of 1st 12 terms ... (2 + 35) * (12 / 2)