Respuesta :
Question:
The sum of first n terms of an AP is given by Sn= 2n square +3n. Find the sixteenth term of the AP
Answer:
16th term of AP is 65
Solution:
Given that,
The sum of first n terms of AP is given by:
[tex]S_n = 2n^2+3n[/tex]
Let nth term in AP be: [tex]a_n[/tex]
Then the nth term is given as:
[tex]a_n = S_n - S_{n-1}[/tex]
Where, from given,
[tex]S_n = 2n^2+3n[/tex]
[tex]\text{For } s_{n-1} \text{ substitute n = n -1 }[/tex]
[tex]S_{n-1} = 2(n-1)^2 + 3(n-1)\\\\S_{n-1} = 2(n^2 -2n + 1}) + 3n - 3\\\\S_{n-1} = 2n^2 -4n + 2 + 3n - 3\\\\S_{n-1} = 2n^2 -n -1[/tex]
Thus nth term is given as:
[tex]a_n = S_n - S_{n-1}\\\\a_n = 2n^2 + 3n - (2n^2-n-1)\\\\a_n = 2n^2 + 3n - 2n^2 +n+1\\\\a_n = 4n + 1[/tex]
To find the 16th term, substitute n = 16
[tex]a_{16} = 4(16) + 1\\\\a_{16} = 64 + 1\\\\a_{16} = 65[/tex]
Thus 16th term of AP is 65
