Let \(\displaystyle{ < }{a}_{{n}}{>}\) and \(\displaystyle{

Octavio Chen

Octavio Chen

Answered question

2022-03-24

Let <an> and <bn> be the arithmetic progression sequences each with common difference 2 such that a1<b1 and let cn={k=1}nak and dn={k=1}nbk. Suppose the points An(an,cn) and Bn(bn,dn) are all lying on the parabola, y=px2+qx+r, then find value of p, q.

Answer & Explanation

Matronola3zw6

Matronola3zw6

Beginner2022-03-25Added 10 answers

The points An(an,cn) and Bn(dn,dn) lie on the parabola. This means that
cn=pan2+qan+r

dn=pbn2+qbn+r
for each n. Subtract the two equations to obtain
dncn=p(bnan)(bn+an)+q(bnan)
Note that bnan=(bn1+2)(an1+2)=bn1an1==b1a1
and dncn=(bnan)+(bn1an1)++(b1a1)=n(b1a1 
so we have n(b1a1)=p(b1a1)(bn+an)+q(b1a1)

n=p(bn+an)+q
where we divided by b1a10. This holds for each n so you can take n=1 and n=2 to obtain
1=p(a1+b1)+q
2=p(a2+b2)+q
Then subtract the two equations (note that a2a1=2 and b2b1=2), and we have 4p=1p=14. To find q, use that (a1,a1) and (a2,a1+a2) lie on the parabola:
a1=p(a1)2+qa1+ra1+a2=p(a2)2+qa2+r
Again, subtract the two equations, substitute a2=a1+2, and then plug in p=14. You'll arrive at 2q=1q=12.

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