The sum of the first seventeen terms of an arithmetic series is 493 The third term is 11. Find the first term and the seventeenth term.
The n-th term is 3*n + 2
To find the first term and seventeenth term of the arithmetic series, we need to apply the formula for finding the sum of an arithmetic series.
The formula for the sum of an arithmetic series is: Sn = (n/2)(a1 + an), where Sn represents the sum of the series, n represents the number of terms, a1 represents the first term, and an represents the last term.
The sum of the first seventeen terms is given as 493, which means Sn = 493 and n = 17. We can plug these values into the formula to get:
493 = (17/2)(a1 + an)
Now, we need to find the values of a1 (first term) and an (seventeenth term). We are also given that the third term is 11, which means a3 = 11.
We can substitute these values into the formula for finding the n-th term of an arithmetic series, which is: an = a1 + (n - 1)d, where d is the common difference.
Substituting the values, we have:
a3 = a1 + (3 - 1)d
11 = a1 + 2d
Since we know the value of a3, we have an equation with two unknowns (a1 and d) and we need to solve it.
By manipulating the equation, we can express d in terms of a1:
11 - a1 = 2d
d = (11 - a1)/2
Substituting this back into the formula for the sum of the series, we have:
493 = (17/2)(a1 + a1 + (17 - 1)((11 - a1)/2))
Now, we can solve this equation for a1.
493 = (17/2)(2a1 + 16(11 - a1)/2)
493 = 17(a1 + 88 - 8a1)/2
493 = 17(88 - 7a1)/2
2 * 493 = 17(88 - 7a1)
286 = 88 - 7a1
7a1 = 88 - 286
7a1 = -198
a1 = -198/7
a1 = -28
Now that we have the value of a1, we can find the seventeenth term by substituting it back into the formula for the n-th term:
an = a1 + (17 - 1)d
an = -28 + 16d
We already found that d = (11 - a1)/2:
d = (11 - (-28))/2
d = 39/2
Now, substitute d into the equation for the seventeenth term:
an = -28 + 16(39/2)
an = -28 + 624/2
an = -28 + 312
an = 284
Therefore, the first term is -28 and the seventeenth term is 284.