Sequence & Series (WS-1)
Question No: 1 [ Maximum Mark: 7 ] (Do not use GDC)
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| Answer : (a) \(7,3,-1,-5 \) (b) \( u_{20}=69\) (c) \( n=27\) | GDC Support | |||||||||||||||||||
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(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
\( \begin{aligned}
u_{20} & =11-4(20) \\
& =11-80 \\
U_{20} & =69
\end{aligned}\)
(c) Given \( u_n=-97\)
\( \begin{aligned}
11-4 n & =-97 \\
-4 n & =-97-11 \\
-4 n & =-108 \\
n & =27
\end{aligned}\)
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Question No: 2 [ Maximum Mark: 7 ] (Do not use GDC)
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| Answer : (a) Proof (b) \( d=5\) (c) \(20th \)term (d) \(S_{20}=1090 \) | GDC Support | |||||||||||||||||||||||
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(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
\( \begin{aligned}
d & =u_2-u 1 \\
& =12-1 \\
d & =5
\end{aligned}\)
(c) we have \( u_1=7 \) and \( d=5\)
\( \begin{aligned}
u_n & =7+(n-1) 5 \\
& =7+5 n-5 \\
u_n & =5 n+2
\end{aligned}\)
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,
\( \begin{aligned}
S_{20} & =\frac{20}{2}[7+102] \\
& =10[109] \\
S_{20} & =1090
\end{aligned}\)
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Question No: 3 [ Maximum Mark: 7 ] (Do not use GDC)
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| Answer : (a) Proof. (b) \( 18,22,26 \text {. }\) (c) \( S_{10}=200 \) | GDC Support | |||||||||||||||||||
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(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 \( 18,22,26\)
(c) Sum up to the first 10 terms of the sequence
\( \begin{aligned}
S_n & =\frac{n}{2}[241+(n-1) d] \\
S_{10} & =\frac{10}{2}[2(2)+(10-1) \times 4] \\
& =5[4+36] \\
& =5 \times 40 \\
S_{10} & =200
\end{aligned}\)
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Question No: 4 [ Maximum Mark: 7 ] (Do not use GDC)
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| Answer : (a) \(u_{10}=2560 \) (b) \( n=8\) (c) \( S_7=635 \) | GDC Support | |||||||||||||||||||
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\(\begin{aligned}
& u_n=u_1 \times r^{n-1} \\
& u_{10}=u_1 \times r^{10-1} \\
& u_{10}=5 \times 2^{10-1} \\
& u_{10}=5 \times 2^9 \\
& u_{10}=5 \times 512 \\
& u_{10}=2560
\end{aligned}\)
(b) given \( u_n=640\)
\( \begin{aligned}
u_n & =640 \\
5 \times 2^{n-1} & =640 \\
2^{n-1} & =\frac{640}{5} \\
2^{n-1} & =128 \\
2^{n-1} & =2^7 \\
n-1 & =7 \\
n & =8
\end{aligned}\)
(c) Substitute values in \( u_1=5\) , \( r=2 \) and \(n=7 \) in \( S_n=\frac{u_1\left[r^n-1\right]}{r-1}\)
\( \begin{aligned}
& S_7=\frac{5\left[2^7-1\right]}{2-1} \\
& S_7=5[128-1] \\
& S_7=5 \times 127 \\
& S_7=635
\end{aligned}\)
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Question No: 5 [ Maximum Mark: 7 ] (Do not use GDC)
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| Answer : (a) \(d=-4 \) (b) \( 18,22,26 \text {. }\) (c) \( S_{20}=-170 \) | GDC Support | |||||||||||||||||||
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(a) \(d=-4 \)
(b) Given that \( u_n=-37\)
\(\begin{aligned}
41+(n-1) d & =-37 \\
20+(n-1) x-3 & =-37 \\
-3(n-1) & =-57 \\
n-1 & =19 \\
n & =20
\end{aligned} \)
(c) Substitute n=20 in \( S_n=\frac{n}{2}\left[u_1+u_n\right]\)
\(\begin{aligned}
S_n & =\frac{n}{2}\left[u_1+u_n\right] \\
S_{20} & =\frac{20}{2}[20+(-37)] \\
& =10[20-37] \\
& =10 \times-17 \\
& =-170
\end{aligned} \)
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