1)The first term of an arithmetic progression is 2.The sum of the first three numbers is 8 terms of the AP is 156.
i)Find the common difference of the AP
ii)Given that the sum of the first n terms of the AP is 416,find n
iii)The third fifth and eighth terms of another AP form the first three terms of a geometric progression.If the common difference of the AP is 3,find;
a)The first term of the GP
b)The sum of the first 9 terms of the GP to 4 significant figures
2)In a figure,the centre of a circle PQ and PR are the tangents.Angle PQS=40° and angle PRS=30°.Giving reasons find;
i)angle RTQ
ii)Angle ORQ
iii)angle RPQ
iv)angle RPQ
3)A figure represents a right angled pyramid with vertix V and a rectangular base PQRS.VP=VQ=VR=VS=18cm.PQ=16cm and QR=12cm.M and O are the mid points of QR and PR respectively. Determine;
a)The length of the projection of line VP on the plane PQRS
b)The size of the angle between line VP and the plane PQRS
c)The size of the angle between the planes VQR and PQRS
1)
i) Let the common difference be d.
The sum of the first three numbers is:
2 + (2+d) + (2+2d) = 8
6 + 3d = 8
3d = 2
d = 2/3
ii) The formula for the sum of the first n terms of an AP is: S_n = n/2 (2a + (n-1)d)
Given that S_8 = 156, we can solve for n:
156 = n/2 (2(2) + (n-1)(2/3))
156 = n(4 + (2/3)(n-1))
156 = 4n + 2n(n-1)/3
156 = 4n + 2n^2 - 2n
2n^2 + 2n - 156 = 0
n^2 + n - 78 = 0
(n+13)(n-12) = 0
n = 12 (as n cannot be negative)
iii)
a) The third term = 2 + 2d = 2 + 2(3) = 8
The fifth term = 2 + 4d = 2 + 4(3) = 14
The eigth term = 2 + 7d = 2 + 7(3) = 23
This forms the GP 8, 14, 23
The common ratio of a GP is given by dividing the n+1th term by the nth term.
So, 14/8 = 23/14
14^2 = 8*23
196 = 184
196 ≠ 184
Therefore, there seems to be an error in the question provided.
2)
i) angle RTQ = angle PSP (Tangent from an external point is perpendicular to the radius)
So, angle RTQ = 90°
ii) angle ORQ = angle OPR (Tangent at point of contact is perpendicular to the radius)
Therefore, angle ORQ = 90°
iii) angle RPQ = angle RPS = angle RQS (Alternate segment theorem)
So, angle RPQ = 30°
iv) angle RPQ = angle PQS (Angles in the same segment are equal)
So, angle RPQ = 40°
3)
a) Length of the projection of line VP on the plane PQRS = VP*cosine of angle between VP and plane PQRS
= 18cos(45°) = 18*(1/√2) = 9√2 cm
b) Angle between line VP and plane PQRS = 45° (complementary angle to the angle of projection)
c) Angle between the planes VQR and PQRS is same as the angle of projection, which is 45°.