Find the sum of the first 60 terms of the arithmetic sequence whose first term is −15 and whose common difference is 3.
I really need help! please can someone help me I need this asap!!
just use your formula
S60 = 60/2 (2a + 59d)
the 60th term is ... -15 + (3 * 59) = 162
the sum of the terms is
... the average of the first and last terms ... (-15 + 162) / 2
... multiplied by the number of terms ... 60
To find the sum of the first 60 terms of the arithmetic sequence, we can use the formula:
Sn = (n/2)(2a + (n-1)d)
where:
Sn is the sum of the first n terms
n is the number of terms
a is the first term
d is the common difference
In this case, n = 60, a = -15, and d = 3. Substituting these values into the formula:
Sn = (60/2)(2(-15) + (60-1)(3))
= 30(-30 + 177)
= 30(147)
= 4410
Therefore, the sum of the first 60 terms of the arithmetic sequence is 4410.
Of course, I can help you with that! To find the sum of the first 60 terms of an arithmetic sequence, you can use the formula for the sum of an arithmetic series.
The formula is:
Sn = (n/2) * (2a + (n-1)d)
Where:
Sn = the sum of the first n terms of the arithmetic series
a = the first term of the arithmetic series
d = the common difference between consecutive terms
n = the number of terms in the series
In your case:
a = -15 (the first term)
d = 3 (the common difference)
n = 60 (the number of terms)
Now let's substitute the values into the formula and calculate the sum:
Sn = (60/2) * (2 * -15 + (60 - 1) * 3)
Simplifying the expression inside the parentheses:
Sn = 30 * (-30 + 59 * 3)
Sn = 30 * (-30 + 177)
Sn = 30 * 147
Sn = 4,410
Therefore, the sum of the first 60 terms of the arithmetic sequence with a first term of -15 and a common difference of 3 is 4,410.
I hope this helps! Let me know if you have any further questions.