1. a1 = 5, a4 = 15
2. a3 = 94, a6 = 85
2. a3 = 94, a6 = 85
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Note that the n-th term of an arithmetic series, a_n, in terms of its first term, a₁, and its common difference, d, is:
a_n = a₁ + (n - 1)d.
(a) Since a₄ = a₁ + (4 - 1)d = a₁ + 3d, we see that:
15 = 5 + 3d, since a₄ = 15 and a₁ = 5
==> d = 10/3.
Thus:
a_n = 5 + (10/3)(n - 1) = (10/3)n + 5/3.
(b) Since a₃ = a₁ + 2d and a₆ = a₁ + 5d, we see that we have the system:
a₁ + 2d = 94 and a₁ + 5d = 85.
Subtracting the first equation from the second gives:
(a₁ + 5d) - (a₁ + 2d) = 85 - 94
==> 3d = -9
==> d = -3.
Then, from the first equation:
a₁ + 2(-3) = 94 ==> a₁ = 100.
Thus:
a_n = 100 - 3(n - 1) = -3n + 103.
I hope this helps!
a_n = a₁ + (n - 1)d.
(a) Since a₄ = a₁ + (4 - 1)d = a₁ + 3d, we see that:
15 = 5 + 3d, since a₄ = 15 and a₁ = 5
==> d = 10/3.
Thus:
a_n = 5 + (10/3)(n - 1) = (10/3)n + 5/3.
(b) Since a₃ = a₁ + 2d and a₆ = a₁ + 5d, we see that we have the system:
a₁ + 2d = 94 and a₁ + 5d = 85.
Subtracting the first equation from the second gives:
(a₁ + 5d) - (a₁ + 2d) = 85 - 94
==> 3d = -9
==> d = -3.
Then, from the first equation:
a₁ + 2(-3) = 94 ==> a₁ = 100.
Thus:
a_n = 100 - 3(n - 1) = -3n + 103.
I hope this helps!