Respuesta :
Answer:
see explanation
Step-by-step explanation:
the sum to n terms of an arithmetic sequence is
[tex]S_{n}[/tex] = [tex]\frac{n}{2}[/tex][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
[tex]S_{n}[/tex] = [tex]\frac{n}{2}[/tex][(2 × 5) + 2(n - 1) ]
= [tex]\frac{n}{2}[/tex](10 + 2n - 2)
= [tex]\frac{n}{2}[/tex](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
The sum of n terms for the arithmetic series is:
[tex]S_n=\frac{n}{2}[8+2n ][/tex]
The value of n for which the series has sum 165 is 11
The given series is:
5 + 7 + 9 +.......
The first value, a = 5
The common difference, d = 7 - 5
d = 2
The sum of the first n terms of an arithmetic series is given by the formula:
[tex]S_n=\frac{n}{2}[2a+(n-1)d ][/tex]
Substituting a = 5, and d = 2 into the sum of the series, we have:
[tex]S_n=\frac{n}{2}[2(5)+(n-1)(2) ]\\\\S_n=\frac{n}{2}[10+2n-2 ]\\\\S_n=\frac{n}{2}[8+2n ][/tex]
To determine the value of n for which the series has sum 165, substitute [tex]S_n=165[/tex] into the sum above
[tex]S_n=\frac{n}{2}[8+2n ]\\\\165=\frac{n}{2}[8+2n ]\\\\2(165)=n(8+2n)\\\\2n^2+8n-330=0\\\\n^2+4n-165=0\\\\n^2-11n+15n-165=0\\\\n(n-11)+15(n-11)=0\\\\(n-11)(n+15)=0\\\\n=11, n=-15[/tex]
The value of n for which the series has sum 165 is 11
Learn more on Arithmetic Progression here: https://brainly.com/question/24205483
