2022 H2 Mathematics Paper 1 Question 9 Arithmetic and Geometric Progressions (APs, GPs)
Answers d = 5 2 a . {d=\frac{5}{2} a.} d = 2 5 a . (bi)
k = 1 2 . {k=\frac{1}{2}.} k = 2 1 . (bii)
43 128 3 . \frac{43}{128} \sqrt{3}. 128 43 3 . Full solutions
(a) Let the common ratio of the geometric series be
r . {r.} r . a = a a + 2 d = a r a + 14 d = a r 2 \begin{align}
a &= a
\\ a + 2d &= ar
\\ a + 14d &= ar^2
\end{align} a a + 2 d a + 14 d = a = a r = a r 2
Taking
( 2 ) − ( 1 ) , {(2) - (1),} ( 2 ) − ( 1 ) , 2 d = a r − a \begin{equation}
2d = ar -a
\end{equation} 2 d = a r − a
Taking
( 3 ) − ( 1 ) , {(3) - (1),} ( 3 ) − ( 1 ) , 14 d = a r 2 − a \begin{equation}
14d = ar^2 -a
\end{equation} 14 d = a r 2 − a
Considering
( 4 ) {(4)} ( 4 ) and
( 5 ) {(5)} ( 5 ) 7 ( a r − a ) = a r 2 − a a r 2 − 7 a r + 8 a = 0 r 2 − 7 r + 6 = 0 ( r − 1 ) ( r − 6 ) = 0 \begin{gather*}
7(ar-a) = ar^2 - a
\\ ar^2 - 7ar + 8a = 0
\\ r^2 - 7 r + 6 = 0
\\ (r - 1)(r - 6) = 0
\end{gather*} 7 ( a r − a ) = a r 2 − a a r 2 − 7 a r + 8 a = 0 r 2 − 7 r + 6 = 0 ( r − 1 ) ( r − 6 ) = 0
Since
d ≠ 0 , {d \neq 0,} d = 0 , r ≠ 1 {r \neq 1} r = 1 so
r = 6. {r=6.} r = 6.
Substituting
r = 6 {r=6} r = 6 into
( 4 ) , {(4),} ( 4 ) , 2 d = 6 a − a d = 5 2 a ■ \begin{align*}
2d &= 6a - a
\\ d &= \frac{5}{2} a \; \blacksquare
\end{align*} 2 d d = 6 a − a = 2 5 a ■ (bi)
S ∞ = sin θ 1 − ( − cos θ ) = sin θ 1 + cos θ = 2 sin θ 2 cos θ 2 1 + 2 cos 2 θ 2 − 1 = 2 sin θ 2 cos θ 2 2 cos 2 θ 2 = sin θ 2 cos θ 2 = tan 1 2 θ ■ \begin{align*}
S_\infty &= \frac{\sin \theta}{1-(-\cos \theta)}
\\ &= \frac{\sin \theta}{1+\cos \theta}
\\ &= \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{1 + 2 \cos^2 \frac{\theta}{2} - 1}
\\ &= \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}}
\\ &= \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}
\\ &= \tan {\textstyle \frac{1}{2}} \theta \; \blacksquare
\end{align*} S ∞ = 1 − ( − cos θ ) sin θ = 1 + cos θ sin θ = 1 + 2 cos 2 2 θ − 1 2 sin 2 θ cos 2 θ = 2 cos 2 2 θ 2 sin 2 θ cos 2 θ = cos 2 θ sin 2 θ = tan 2 1 θ ■ (bii)
S 7 = sin π 3 ( 1 − ( − cos π 3 ) 7 ) 1 − ( − cos π 3 ) = 3 2 ( 1 − ( − 1 2 ) 7 ) 1 + 1 2 = 43 128 3 ■ \begin{align*}
S_7 &= \frac{\sin {\textstyle \frac{\pi}{3}} \left( 1 - (-\cos {\textstyle \frac{\pi}{3}})^7 \right)}{1-(-\cos {\textstyle \frac{\pi}{3}})}
\\ &= \frac{\frac{\sqrt{3}}{2} \left( 1 - ({\textstyle - \frac{1}{2}} \right)^7 )}{1+\frac{1}{2}}
\\ &= \frac{43}{128} \sqrt{3} \; \blacksquare
\end{align*} S 7 = 1 − ( − cos 3 π ) sin 3 π ( 1 − ( − cos 3 π ) 7 ) = 1 + 2 1 2 3 ( 1 − ( − 2 1 ) 7 ) = 128 43 3 ■ Question Commentary
Part (a) is a repeat of the rather popular algebraic type of AP/GP question,
seen in 2016 paper 1 question 4 and
2007 paper 1 question 10 , where we link the two types together.
One strategy is to subtract the equations that can be formed to isolate the common
difference d , {d,} d , r {r} r
Meanwhile, part (b) tests the sum to infinity formula with a dash of trigonometry
where we will have to use the double angle formula to get the required final form.