>......> ........... answer plz
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Answer:
see explanation
Step-by-step explanation:
the sum to n terms of an arithmetic sequence is
=
[2a + (n - 1)d ]
where d is the common difference and a is the first term
here d = 9 - 7 = 7 - 5 = 2 and a = 5, hence
=
[(2 × 5) + 2(n - 1) ]
= (10 + 2n - 2)
= (2n + 8)
= n² + 4n
When sum = 165, then
n² + 4n = 165 ← rearrange into standard form
n² + 4n - 165 = 0 ← in standard form
(n + 15)(n - 11) = 0 ← in factored form
equate each factor to zero and solve for n
n + 15 = 0 ⇒ n = - 15
n - 11 = 0 ⇒ n = 11
but n > 0 ⇒ n = 11