For an arithmetic sequence, a_{19} = 78. If the common difference is -3, find
a_1 = (i found it to be 132)
the sum of the first 57 terms =??
-I know there is a formula but i cant think of what it is
Sumn = (n/2)[2a + n-1)d]
Sumn = (n/2)[2a + (n-1)d]
To find a_1 (the first term) of an arithmetic sequence, you can use the formula:
a_n = a_1 + (n - 1)d
where a_n is the nth term, a_1 is the first term, n is the number of terms, and d is the common difference.
In this case, you know that a_19 = 78, and the common difference, d, is -3. Substituting these values into the formula, you get:
78 = a_1 + (19 - 1)(-3)
78 = a_1 + 18(-3)
78 = a_1 - 54
Adding 54 to both sides, you get:
132 = a_1
So, you are correct that a_1 = 132.
Now, to find the sum of the first 57 terms, you can use the formula for the sum of an arithmetic series:
S_n = (n/2)(a_1 + a_n)
where S_n is the sum of the first n terms.
Substituting the given values:
n = 57
a_1 = 132 (which you found correctly)
a_n = a_1 + (n - 1)d
Since a_n = a_1 + (n - 1)d, we can substitute it into the sum formula:
S_n = (n/2)(a_1 + a_1 + (n - 1)d)
S_n = (n/2)(2a_1 + (n - 1)d)
S_57 = (57/2)(2(132) + (57 - 1)(-3))
Simplifying the expression further:
S_57 = (57/2)(264 - 168)
S_57 = (57/2)(96)
S_57 = 57(48)
S_57 = 2736
Therefore, the sum of the first 57 terms is 2736.