Sequence & Series (WS-1)

Question No: 1 [ Maximum Mark: 7 ] (Do not use GDC)

The general term of an arithmetic sequence is $ u_n=11-4 n$
(a)	Find the first four terms of the sequence.	[2 marks]
(b)	Find the 2oth term of the sequence	[2 marks]
(c)	Find the nth term of the sequence is -97 find ‘n’	[2 marks]

Formula Sheet

Answer : (a) $7,3,-1,-5 $ (b) $ u_{20}=69$ (c) $ n=27$

GDC Support

(a) Given $ u_n=11-4 n$

$ \begin{aligned}
& u_1=11-4(1)=7 \\
& u_2=11-4(2)=3 \\
& u_3=11-4(3)=-1 \\
& u_4=11-4(4)=-5
\end{aligned}$

Therefore the first four terms of the sequence are $7,3,-1,-5 $

(b) Substitute n = 20 in $ u_n=11-4 n$ we get

$ \begin{aligned}
u_{20} & =11-4(20) \\
& =11-80 \\
U_{20} & =69
\end{aligned}$

$ \begin{aligned}
11-4 n & =-97 \\
-4 n & =-97-11 \\
-4 n & =-108 \\
n & =27
\end{aligned}$

Video

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Question No: 2 [ Maximum Mark: 7 ] (Do not use GDC)

The first three terms of an arithmetic sequence are $ u_1, 3 u_1-9,2 u_1+3 $
(a)	Show that $ u_1=7 $	[3 marks]
(b)	Find the common difference d	[2 marks]
(c)	Which term of the sequence is 102.	[2 marks]
(d)	Find the sum of the first 20 terms of the sequence	[3 marks]

Formula Sheet

Answer : (a) Proof (b) $ d=5$ (c) $20th $term (d) $S_{20}=1090 $

GDC Support

(a) Given, the first three terms are $ u_1, 3 u_1-9,2 u_1+3 $

$ \begin{aligned}
3 u_1-9-u_1 & =2 u_1+3-3 u_n+9 \\
2 u_1-q & =-u_1+12 \\
3 u_l & =21 \\
u_1 & =7
\end{aligned}$

(b) Common difference $ d=u_2-u 1$

$ \begin{aligned}
u_2 & =3 u_1-9 \\
& =3(7)-9 \\
& =21-9 \\
u_2 & =12
\end{aligned}$

We have the first term is $ u_1=7 $

therefore the common difference d is

$ \begin{aligned}
d & =u_2-u 1 \\
& =12-1 \\
d & =5
\end{aligned}$

$ \begin{aligned}
u_n & =7+(n-1) 5 \\
& =7+5 n-5 \\
u_n & =5 n+2
\end{aligned}$

Given $ u_n=1o2$, have to find $ n $

$ \begin{aligned}
5 n+2 & =102 \\
5 n & =100 \\
n & =20
\end{aligned}$

(d) To find $ S_{20}$ substitute n=20 in $ S_{n}=\frac{n}{2}\left[u_1+u_n\right]$

we get,

$ \begin{aligned}
S_{20} & =\frac{20}{2}[7+102] \\
& =10[109] \\
S_{20} & =1090
\end{aligned}$

Video

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Question No: 3 [ Maximum Mark: 7 ] (Do not use GDC)

Consider the sequence $ 2,6,10,14, \ldots \ldots$
(a)	Show that it’s an arithmetic sequence.	[2 marks]
(b)	Find the next three terms of the sequence	[2 marks]
(c)	Find the sum of up to 10 terms of the sequence	[2 marks]

Formula Sheet

Answer : (a) Proof. (b) $ 18,22,26 \text {. }$ (c) $ S_{10}=200 $

GDC Support

(a) Given sequence is $ 2,6,10,14, \ldots \ldots$

$ \begin{aligned}
& d_1=6-2=4 \\
& d_2=10-6=4 \\
& d_3=14-10=4
\end{aligned}$

Since $ d_1=d_2=d_3 $

(b) Next three terms are $ 18,22,26$

$ \begin{aligned}
S_n & =\frac{n}{2}[241+(n-1) d] \\
S_{10} & =\frac{10}{2}[2(2)+(10-1) \times 4] \\
& =5[4+36] \\
& =5 \times 40 \\
S_{10} & =200
\end{aligned}$

Video

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Question No: 4 [ Maximum Mark: 7 ] (Do not use GDC)

A geometric sequence has the first term 5 and a common ratio of 2.
(a)	Find the 1oth term of the sequence.	[2 marks]
(b)	The nth term of the sequence is 640, find $n$	[2 marks]
(c)	Find the sum of the first 7 terms of the sequence	[2 marks]

Formula Sheet

Answer : (a) $u_{10}=2560 $ (b) $ n=8$ (c) $ S_7=635 $

GDC Support

(a) Plugin in $ u_1=5$ and $ r=2$ in $ u_n=u_1 \times r^{n-1}$

$\begin{aligned}
& u_n=u_1 \times r^{n-1} \\
& u_{10}=u_1 \times r^{10-1} \\
& u_{10}=5 \times 2^{10-1} \\
& u_{10}=5 \times 2^9 \\
& u_{10}=5 \times 512 \\
& u_{10}=2560
\end{aligned}$

(b) given $ u_n=640$

$ \begin{aligned}
u_n & =640 \\
5 \times 2^{n-1} & =640 \\
2^{n-1} & =\frac{640}{5} \\
2^{n-1} & =128 \\
2^{n-1} & =2^7 \\
n-1 & =7 \\
n & =8
\end{aligned}$

$ \begin{aligned}
& S_7=\frac{5\left[2^7-1\right]}{2-1} \\
& S_7=5[128-1] \\
& S_7=5 \times 127 \\
& S_7=635
\end{aligned}$

Video

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Question No: 5 [ Maximum Mark: 7 ] (Do not use GDC)

Consider the arithmetic sequence is $ 20,17,14,11, \ldots$
(a)	Find the common difference ‘d’ of the sequence.	[2 marks]
(b)	Given that $ u_n=-37$ find ‘n’	[2 marks]
(c)	Find the sum of up to 20 terms of the sequence	[2 marks]

Formula Sheet

Answer : (a) $d=-4 $ (b) $ 18,22,26 \text {. }$ (c) $ S_{20}=-170 $

GDC Support

(a) $d=-4 $

(b) Given that $ u_n=-37$

$\begin{aligned}
41+(n-1) d & =-37 \\
20+(n-1) x-3 & =-37 \\
-3(n-1) & =-57 \\
n-1 & =19 \\
n & =20
\end{aligned} $

$\begin{aligned}
S_n & =\frac{n}{2}\left[u_1+u_n\right] \\
S_{20} & =\frac{20}{2}[20+(-37)] \\
& =10[20-37] \\
& =10 \times-17 \\
& =-170
\end{aligned} $

Video

Question No: 1 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 2 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 3 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 4 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 5 [ Maximum Mark: 7 ] (Do not use GDC)

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Sequence & Series (WS-1)

Question No: 1 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 2 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 3 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 4 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 5 [ Maximum Mark: 7 ] (Do not use GDC)

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