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## Algebra: Sequence and Series Test-3

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Question 1 |

The sum 9 + 16 + 25 + 36 + â€¦.. + 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 (10

^{12}+ 25)^{2}â€“ (10^{12}â€“25)^{2}=10n, then the value of n is20 | |

14 | |

10 | |

5 |

Question 3 Explanation:

(10

=> (10

=>Â 2 x 10

=>10

=>n=14

^{12}+ 25)^{2}â€“ (10^{12}â€“25)^{2}=10n=> (10

^{12}+ 25+10^{12}â€“25) (10^{12}+ 25-10^{12}+25)=10n=>Â 2 x 10

^{12 }x 2 x25=10^{n}=>10

^{14}=10^{n}=>n=14

Question 4 |

Given 1 + 2 + 3 + 4 + â€¦â€¦.+ 10 = 55, then the sum 6 + 12 + 18 + 24 + â€¦â€¦.+ 60 is equal to:

300 | |

655 | |

330 | |

455 |

Question 4 Explanation:

The given series is = 6 X (1+2+3+â€¦.+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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