Sequence & Series (WS-3)

Question No: 11 [ Maximum Mark: 7 ] (Do not use GDC)

The first three terms of a sequence are \(4,8,16 \)
(a)	show that it is a geometric sequence	[2 marks]
(b)	Find the next three terms	[1 mark]
(c)	Find the 10^th term of the sequence	[2 marks]
(d)	Find the sum of up to 10 terms of the sequence	[2 marks]

Formula Sheet

Answer : (a) Proof (b) \( 32,64,128 \) (c) \(u_{10}=2048 \) (d) \(S_{10}=4092 \)

GDC Support

(a) given first three terms are \( 4,8,16\)

\( \begin{aligned}
& r_1=\frac{8}{4}=2 \\
& r_2=\frac{16}{8}=2
\end{aligned}\)

since \(r_1=r_2 \) the three terms form a geometric sequence with the common ratio \(r=2 \)

(b) The next three terms,

\( \begin{aligned}
& 16 \times 2=32 \\
& 32 \times 2=64 \\
& 64 \times 2=128
\end{aligned}\)

The next three terms are \( 32,64,128\)

\(\begin{aligned}
u_{10} & =4 \times 2^{10-1} \\
& =4 \times 2^9 \\
& =2048
\end{aligned} \)

(d) Substitute n=10 in \(S_n=\frac{u_{1}\left[r^n-1\right]}{r-1} \)

\(\begin{aligned}
& S_n=\frac{41\left[r^n-1\right]}{r-1} \\
& S_{10}=\frac{4\left[2^{10}-1\right]}{2-1} \\
& S_{10}=4092
\end{aligned} \)

Video

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Question No: 12 [ Maximum Mark: 8 ] (Do not use GDC)

Consider the geometric sequence \( -27 + 9 – 3 + . . . . .\)
(a)	Find the 7^th term of the series	[2 marks]
(b)	Find the sum of the first 10 terms of this sequence	[2 marks]
(c)	Show that sum up to the infinity of this series is convergent	[2 marks]
(d)	Find the sum up to the infinity of this series	[2 marks]

Formula Sheet

Answer : (a) \(u_{7}=\frac{-7}{243} \) (b) \(S_{10}=-40.5 \) (c) \( |r|=\left|-\frac{1}{3}\right|<1\) (d) \(\frac{-81}{4}\)

GDC Support

Given series is \(-27+9-3+\cdots \)

here \(u_1=-21 \),

\( \begin{aligned}
& r=\frac{9}{-27} \\
& r=\frac{-1}{3}
\end{aligned}\)

(a) Find the 7^th term of the series

\(\begin{aligned}
u_n & =u_1 \times r`^{n-1} \\
u_7 & =-21 \times\left(\frac{-1}{3}\right)^{7-1} \\
& =\frac{-7}{243}
\end{aligned} \)

(b) Substitute n=10 in \(S_n=\frac{u_{1}\left[r^n-1\right]}{r-1} \)

\( \begin{aligned}
& S_{10}=\frac{-27\left[1-\left(\frac{-1}{3}\right)^{10}\right]}{1-\frac{1}{3}} \\
& S_{10}=-40.5
\end{aligned}\)

Therefore the sum up to the infinity of the series is convergent

(d) Sum up to the infinity of the series

\(\begin{aligned}
S_{\infty} & =\frac{u_1}{1-r} \\
S_{\infty} & =\frac{-27}{1-\left(-\frac{1}{3}\right)} \\
& =\frac{-27}{1+\frac{1}{3}} \\
& =\frac{-81}{4}
\end{aligned} \)

Video

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Question No: 13 [ Maximum Mark: 4 ] (Do not use GDC)

In an arithmetic sequence, the first term is 3 and the 25^th term is 51.
(a)	Find the first term of the common difference of the sequence	[2 marks]
(b)	Find the sum of the first 30 terms of the sequence.	[2 marks]

Formula Sheet

Answer : (a) \( d=2\) (b) \( S_{30}=960\)

GDC Support

Given that the arithmetic sequence has the first term is 3 and the 25^th term is 51.
(a) here \( u_1=3 \) and \( u_{25}=51\)

\( \begin{aligned}
& u_{25}=51 \\
& u_1+(25-1) d=51 \\
& 3+24 d=51 \\
& 24 d=48 \\
& d=\frac{48}{24} \\
& d=2
\end{aligned}\)

\( \)

(b) Sum up to the first 30 terms of the sequence,

\(\begin{aligned}
S_n & =\frac{n}{2}\left[2 u_1+(n-1) d\right] \\
S_{30} & =\frac{30}{2}[2 \times 3+(30-1) \times 2] \\
S_{30} & =15[6+58] \\
& =15 \times 64 \\
S_{30} & =960
\end{aligned} \)

Video

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Question No: 14 [Maximum Mark: 9] (Do not use GDC)

A geometric series has a second term of -10 and a 5th term as 80
(a)	Find the common ratio ‘r’	[2 marks]
(b)	Find the first term	[2 marks]
(c)	The nth term of the series is -2560 find ‘n’	[2 marks]
(d)	The sum up to ‘k’ terms of this series is -6825, find ‘k’	[3 marks]

Formula Sheet

Answer : (a) \( r=-2\) (b) \( u_1=5\) (c) \( n=10\) (d) \( k=12\)

GDC Support

Given the geometric series has a second term of -10 and a 5th term as 80(a) here \( U_2=-10\) and \( u_5=80\)

\( u_2=-10 \Longrightarrow u_1 \times r=-10\) and

\(u_5=80 \Rightarrow u_1 \times r^4=80 \)

Divide the two-equation to get the common ratio ‘r’

\( \begin{aligned}
\frac{u_{1} \times r^4}{u_{1} \times r} & =\frac{80}{-10} \\
r^3 & =-8 \\
r & =-2
\end{aligned}\)

(b) First term

\(\begin{aligned}
41 \times r & =-10 \\
u_1 & =\frac{-10}{-2} \\
u_1 & =5
\end{aligned} \)

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

(d) The sum up to ‘k’ terms of this series is -6825, to find ‘k’

\(\begin{aligned}
& S_k=-6825 \\
& \frac{41\left[r^k-1\right]}{r-1}=-6825 \\
& \frac{5\left[(-2)^k-1\right]}{-2-1}=-6825 \\
& (-2)^k-1=\frac{-6825 x-3}{5} \\
& (-2)^k=4095+1
\end{aligned} \)

\(\begin{aligned}
(-2)^k & =(-2)^{12} \\
k & =12
\end{aligned} \)

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Question No: 15 [ Maximum Mark: 4 ] (GDC required)

In an arithmetic sequence \(u_{14}=-52 \) and \(u_{40}=-156 \).
(a)	Find the Common difference ‘d’	[2 marks]
(b)	Find the first term \(u_{1} \)
(c)	Find the 50th term of the series \( \)	[2 marks]

Formula Sheet

Answer : (a) \(d=-4 \) (b) \(u_{1}=0 \) (c) \(u_{50}=-4900 \)

GDC Support

(a) Given the arithmetic sequence has \(u_{14}=-52 \) and \(u_{40}=-156 \).

\( \begin{gathered}
u_{14}=-52 \\
u_{11}+(14-1) d=-52 \\
u_1+13 d=-52 \\
u_{40}=-156 \\
u_1+39 d=-156
\end{gathered}\)

solving the two simultaneous equations we get

\( \begin{aligned}
41+39 d & =-156 \\
-41+13 d & =-52 \\
\hline 26 d & =-104 \\
d & =-\frac{104}{26} \\
d & =-4
\end{aligned}\)

(b) First term

\( \begin{gathered}
u_1+13 d=-52 \\
u_1+13(-4)=-52 \\
u_1=-52+52 \\
u_1=0
\end{gathered}\)

\( \begin{aligned}
& u_n=\frac{n}{2}\left[2 u_1+(n-1) d\right] \\
& u_{50}=\frac{50}{2}[2 \times 0+(50-1) x-4] \\
& u_{50}=25[49 \times-4] \\
& u_{50}=-4900
\end{aligned}\)

Video

Question No: 11 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 12 [ Maximum Mark: 8 ] (Do not use GDC)

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Question No: 13 [ Maximum Mark: 4 ] (Do not use GDC)

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Question No: 14 [Maximum Mark: 9] (Do not use GDC)

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Question No: 15 [ Maximum Mark: 4 ] (GDC required)

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Sequence & Series (WS-3)

Question No: 11 [ Maximum Mark: 7 ] (Do not use GDC)

..

Question No: 12 [ Maximum Mark: 8 ] (Do not use GDC)

.

Question No: 13 [ Maximum Mark: 4 ] (Do not use GDC)

.

Question No: 14 [Maximum Mark: 9] (Do not use GDC)

.

Question No: 15 [ Maximum Mark: 4 ] (GDC required)

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Courses

General

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