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## Algebra: Sequence and Series Test-3

Congratulations - you have completed Algebra: Sequence and Series Test-3. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%%
 Question 1
The sum 9 + 16 + 25 + 36 + ….. + 100 is equal to:
 A 350 B 380 C 400 D 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
$\begin{array}{l}=\frac{10(11)(21)}{6}-5\\=385-5\\=380\end{array}$
 Question 2
Find the sum of the first five terms of the following series.
$\displaystyle \frac{1}{1\times 4}+\frac{1}{4\times 7}+\frac{1}{7\times 10}+..........+$
 A 9/32 B 7/16 C 5/16 D 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
 A 20 B 14 C 10 D 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
 Question 4
Given 1 + 2 + 3 + 4 + …….+ 10 = 55, then the sum 6 + 12 + 18 + 24 + …….+ 60 is equal to:
 A 300 B 655 C 330 D 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
$\displaystyle \left( 1-\frac{1}{2} \right)\,\left( 1-\frac{1}{3} \right)\,\left( 1-\frac{1}{4} \right)\,......\left( 1-\frac{1}{n} \right)\,\,\,\,gives:$
 A 1/n B 2/n C {2(n-1)}/n D 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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