Question

In an AP, d=1.6, a=6
\[S_{n}=pn^2+qn \]

Find p and q

Answer 1

@experimentX @lgbasallote @Callisto Try this? :)

Answer 2

I'm going to give wrong answers :(
T(n) = 6+1.6(n-1)
S(n) = [6+ ( 6+1.6(n-1)) ] n/2
= (12+1.6n -1.6) n/2
= (1.6n^2 + 10.4n)/2
= 0.8n^2 + 5.2n
p= 0.8, q=5.2

Answer 3

a is first term right?

A_n = 6 +(n-1)1.6
A_n = 6 + 1.6n - 1.6
A_n = 4.4 + 1.6n

S_n = n(A_1 + A_n)/2

pn^2 + qn = n(6 + 4.4 + 1.6n)/2
n(pn + q) = n(10.4 + 1.6n)/2

pn + q = 5.2 + 0.8n

Answer 4

lol i didnt know what to do with that...genius caLLISTO!

Answer 5

@Callisto is correct :D

Answer 6

Oh... I must be dreaming :(....