The 9th term of an arithmetic series is 5/2
1 answer:
Answer:
- 8050
Step-by-step explanation:
The n th term of an arithmetic sequence is
= a + (n - 1)d
where a is the first term and d the common difference.
We require to find both a and d
Given the 9 th term is 2.5 , then
a + 8d = 2.5 → (1)
Given the sum of the second and fifth term is 27, then
a + d + a + 4d = 27, that is
2a + 5d = 27 → (2)
Multiply (1) by - 2 and add to (2) to eliminate a
- 2a - 16d = - 5 → (3)
Add (2) and (3) term by term
- 11d = 22 ( divide both sides by - 11 )
d = - 2
Substitute d = - 2 into (1) and solve for a
a - 16 = 2.5 ( add 16 to both sides )
a = 18.5
The sum to n terms of an arithmetic sequence is
=
[ 2a + (n - 1)d ], thus
= 50 [ (2 × 18.5) + (99 × - 2) ] = 50(37 - 198) = 50(- 161) = - 8050
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