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Algebra: Sequence and Series Test-3
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Question 1 |
The sum 9 + 16 + 25 + 36 + ….. + 100 is equal to:
350 | |
380 | |
400 | |
420 |
Question 1 Explanation:
Let p =sum of square of first 10 natural numbers
q= sum of square of 1st 2 natural numbers. Thus the given series is p-q
Thus the series
$ \begin{array}{l}=\frac{10(11)(21)}{6}-5\\=385-5\\=380\end{array}$
q= sum of square of 1st 2 natural numbers. Thus the given series is p-q
Thus the series
$ \begin{array}{l}=\frac{10(11)(21)}{6}-5\\=385-5\\=380\end{array}$
Question 2 |
Find the sum of the first five terms of the following series.
$ \displaystyle \frac{1}{1\times 4}+\frac{1}{4\times 7}+\frac{1}{7\times 10}+..........+$
$ \displaystyle \frac{1}{1\times 4}+\frac{1}{4\times 7}+\frac{1}{7\times 10}+..........+$
9/32 | |
7/16 | |
5/16 | |
1/210 |
Question 2 Explanation:
$ \begin{array}{l}\frac{1}{1\times 4}+\frac{1}{4\times 7}+\frac{1}{7\times 10}+..........\\=\frac{1}{3}\left( \begin{array}{l}1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}\\-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\end{array} \right)\\=\frac{15}{16}\times \frac{1}{3}=\frac{5}{16}\end{array}$
Question 3 |
If (1012 + 25)2 – (1012 –25)2 =10n, then the value of n is
20 | |
14 | |
10 | |
5 |
Question 3 Explanation:
(1012 + 25)2 – (1012 –25)2 =10n
=> (1012 + 25+1012 –25) (1012 + 25-1012 +25)=10n
=> 2 x 1012 x 2 x25=10n
=>1014=10n
=>n=14
=> (1012 + 25+1012 –25) (1012 + 25-1012 +25)=10n
=> 2 x 1012 x 2 x25=10n
=>1014=10n
=>n=14
Question 4 |
Given 1 + 2 + 3 + 4 + …….+ 10 = 55, then the sum 6 + 12 + 18 + 24 + …….+ 60 is equal to:
300 | |
655 | |
330 | |
455 |
Question 4 Explanation:
The given series is = 6 X (1+2+3+….+9+10)= 6 X 55 = 330.
Question 5 |
When simplified the product
$\displaystyle \left( 1-\frac{1}{2} \right)\,\left( 1-\frac{1}{3} \right)\,\left( 1-\frac{1}{4} \right)\,......\left( 1-\frac{1}{n} \right)\,\,\,\,gives:$
$\displaystyle \left( 1-\frac{1}{2} \right)\,\left( 1-\frac{1}{3} \right)\,\left( 1-\frac{1}{4} \right)\,......\left( 1-\frac{1}{n} \right)\,\,\,\,gives:$
1/n | |
2/n | |
{2(n-1)}/n | |
2/{n(n-1)} |
Question 5 Explanation:
$ \displaystyle \begin{array}{l}\left( 1-\frac{1}{2} \right)\,\left( 1-\frac{1}{3} \right)\,\left( 1-\frac{1}{4} \right)\,......\left( 1-\frac{1}{n} \right)\,\,\,\,\\=\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times .......\times \frac{n-1}{n}\\=\frac{1}{n}\end{array}$
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