Answer:
Sāā = 455
Step-by-step explanation:
the sum to n terms of an arithmetic is
[tex]S_{n}[/tex] = [tex]\frac{n}{2}[/tex] [ 2aā + (n - 1)d ]
where aā is the first term and d the common difference
We require to find both aā and d
The nth term of an arithmetic sequence is
[tex]a_{n}[/tex] = aā + (n - 1)d
given aā = 20 and aāā = 65 , then
aā + 3d = 20 ā (1)
aā + 12d = 65 ā (2)
subtract (1) from (2) term by term to eliminate aā
9d = 45 ( divide both sides by 9 )
d = 5
substitute d = 5 into (1) and solve for aā
aā + 3(5) = 20
aā + 15 = 20 ( subtract 15 from both sides )
aā = 5
Then
Sāā = [tex]\frac{13}{2}[/tex] [ (2 Ć 5 ) + (12 Ć 5 ) }
= 6.5 (10 + 60)
= 6.5 Ć 70
= 455
