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Algebra: Sequence and Series Test-1
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Question 1 |
The next number of the sequence
3, 5, 9, 17, 33 ……..is:
3, 5, 9, 17, 33 ……..is:
65 | |
60 | |
50 | |
49 |
Question 1 Explanation:
The sequence is 2n+1
Thus the next term is 6th term.
2n+1 = 26+1 = 64+1=65.
The correct option is (a)
Question 2 |
The next term of the sequence$ \displaystyle \frac{1}{2},\,\,3\frac{1}{4},\,\,6,\,\,8\frac{3}{4}........is:$
$ \displaystyle 10\frac{1}{4}$ | |
$ \displaystyle 10\frac{3}{4}$ | |
$ \displaystyle 11\frac{1}{4}$ | |
$ \displaystyle 11\frac{1}{2}$ |
Question 2 Explanation:
The general term is
$ \frac{2+11(n-1)}{4}$
The 5th term will be
$latex \frac{46}{4}=\frac{23}{2}=11\frac{1}{2}$
$ \frac{2+11(n-1)}{4}$
The 5th term will be
$latex \frac{46}{4}=\frac{23}{2}=11\frac{1}{2}$
Question 3 |
Find the missing number of the sequence:
“3, 14, 25, 36, 47, ? ”
“3, 14, 25, 36, 47, ? ”
1114 | |
1111 | |
1113 | |
None of these |
Question 3 Explanation:
The general term is 3+11(n-1).
The 6th term is 58.
The 6th term is 58.
Question 4 |
The sum (101+ 102 + 103 + ….. + 200) is equal to:
15000 | |
15025 | |
15050 | |
25000 |
Question 4 Explanation:
The given expression = total of all natural numbers till
200- total of all natural numbers till 100.
$ \begin{array}{l}=\frac{200(200+1)}{2}-\frac{100(100+1)}{2}\\=20100-5050\\=15050\end{array}$
$ \begin{array}{l}=\frac{200(200+1)}{2}-\frac{100(100+1)}{2}\\=20100-5050\\=15050\end{array}$
Question 5 |
Which term of the series 72, 63, 54………..is zero?
11th | |
10th | |
9th | |
8th |
Question 5 Explanation:
The general term is 72-9(n-1).
Thus,$ \displaystyle \begin{array}{l}72-9\left( n-1 \right)=0\\=>72=9(n-1)\\=>n-1=8\\=>n=9\end{array}$.
Thus,$ \displaystyle \begin{array}{l}72-9\left( n-1 \right)=0\\=>72=9(n-1)\\=>n-1=8\\=>n=9\end{array}$.
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