GR12: 2021 TEST 2
TOTAL: 60
SEQUENCES & SERIES AND GEOMETRY
TIME: 70mins
1. Given the sequence 8; 3; −2; … determine
1.1 the 𝑛𝑡ℎ term of the sequence.
1.2 𝑇20
1.3 the sum of the first 20 terms.
(2)
(2)
(2)
2. Evaluate:
(3)
7
∑ 3 ∙ 2𝑛−1
𝑛=3
3. The first 2 terms of a geometric series are 64 and 16.
If the last term is
1
, determine how many terms are in the series.
64
(4)
4. The difference between the 8th term and the 3rd term of a geometric
sequence is equal to 31 times the 3rd term. Determine the value of the
constant ratio 𝑟.
(5)
5. Determine the largest number of terms for which
𝑛
∑ (2𝑘 − 3) < 48
𝑘=1
(6)
6. If 1 + log 4 𝑥 ; 𝑎 + 2 log 4 𝑥 ; 𝑏 + 3 log 4 𝑥 is an arithmetic sequence,
then 𝑎 = ⋯ (write down the letter corresponding to the answer)
A.
1+𝑏
2
B. 𝑏 − 𝑎
C.
5 log4 𝑥
2
D. 𝑏 − 1
(2)
7. The radius of the first circle of an infinite sequence of circles is 7𝑐𝑚 and
1
the radius of each circle is of the radius of its predecessor. Calculate
8
the sum of the areas of all the circles in the sequence.
(4)
8. ANSWER THIS QUESTION ON THE ANSWER SHEET PROVIDED.
In the diagram below, points D and E lie on sides AB and AC of ∆𝐴𝐵𝐶
respectively such that DE//AC. DC and BE are joined.
Given below is the partially completed proof of the theorem that states
that if in any ∆𝐴𝐵𝐶, DE is parallel to BC, then
𝐴𝐷
𝐷𝐵
=
Complete the proof by filling in the missing parts.
Construction: construct altitudes ℎ and 𝑘 in ∆𝐴𝐷𝐸.
1
𝑎𝑟𝑒𝑎 ∆𝐴𝐷𝐸 2 (𝐴𝐷)ℎ
… ..
=
=
𝑎𝑟𝑒𝑎 ∆𝐷𝐸𝐵 1 (𝐵𝐷)ℎ
….
2
𝑎𝑟𝑒𝑎 ∆𝐴𝐷𝐸
… ..
𝐴𝐸
=
=
𝑎𝑟𝑒𝑎 ∆𝐷𝐸𝐶
….
𝐸𝐶
But 𝑎𝑟𝑒𝑎 ∆𝐷𝐸𝐵 = ………………………. (reason……………)
∴
𝑎𝑟𝑒𝑎 ∆𝐴𝐷𝐸
𝑎𝑟𝑒𝑎 ∆𝐷𝐸𝐵
∴
= …………………..
𝐴𝐷 𝐴𝐸
=
𝐷𝐵 𝐸𝐶
𝐴𝐸
𝐸𝐶
.
(5)
9.
ANSWER THIS QUESTION ON THE ANSWER SHEET PROVIDED.
In the sketch below P, Q, N and M lie on the circle.
KL//PM and LM = MN
P
K
1
2
4Q
3
1
4
L
2R
3
1
2
/
2 1
M
/
N
9.1
Prove that ∆𝑁𝑄𝐿|||∆𝑃𝑀𝐿 .
(4)
9.2
Hence, prove that 𝑄𝐿. 𝑃𝐿 = 2𝑁𝑀 2 .
(5)
P.t.o…….
10. ANSWER THIS QUESTION ON THE ANSWER SHEET PROVIDED.
In the figure below A, B, C and E are points on a circle. AE bisects 𝐵𝐴̂𝐶
and BC and AE intersect at D.
10.1 Prove that ∆𝐴𝐵𝐷|||∆𝐴𝐸𝐶
(4)
10.2 Prove that 𝐴𝐵. 𝐴𝐶 = 𝐴𝐷. 𝐴𝐸
(2)
10.3 Prove that ∆𝐵𝐴𝐷|||∆𝐸𝐶𝐷
(4)
10.4 Prove that
𝐵𝐷
𝐸𝐷
=
𝐴𝐷
𝐶𝐷
10.5 Hence, show that 𝐴𝐵. 𝐴𝐶 = 𝐴𝐷2 + 𝐵𝐷. 𝐷𝐶
(1)
(5)
ANSWER SHEET FOR QUESTIONS 8 – 10
QUESTION 8:
Construction: construct altitudes ℎ and 𝑘 in ∆𝐴𝐷𝐸.
1
𝑎𝑟𝑒𝑎 ∆𝐴𝐷𝐸 2 (𝐴𝐷)ℎ
=
=
𝑎𝑟𝑒𝑎 ∆𝐷𝐸𝐵 1 (𝐵𝐷)ℎ
2
𝑎𝑟𝑒𝑎 ∆𝐴𝐷𝐸
𝐴𝐸
=
=
𝑎𝑟𝑒𝑎 ∆𝐷𝐸𝐶
𝐸𝐶
But 𝑎𝑟𝑒𝑎 ∆𝐷𝐸𝐵 = ………………………. (
∴
∴
𝑎𝑟𝑒𝑎 ∆𝐴𝐷𝐸
𝑎𝑟𝑒𝑎 ∆𝐷𝐸𝐵
𝐴𝐷
𝐷𝐵
=
)
= …………………..
𝐴𝐸
𝐸𝐶
(5)
QUESTION TEN:
SOLUTION
10.1
MARKS
4
10.2
2
10.3
4
10.4
1
10.5
5
QUESTION NINE
P
K
2
1
3
4Q
1
4
2R
3
1
L
2
/
2
M
1
/
N
SOLUTION
9.1
MARKS
4
9.2
5