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## Algebra: Sequence and Series Test-4

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Question 1 |

If the 4

^{th}term of an arithmetic progression is 14 and the 12^{th}term is 70, then the first, term is:â€“10 | |

â€“7 | |

+7 | |

+10 |

Question 1 Explanation:

Let the common difference be d.

14+8d=70,

=> d=7.

First term is 14-3d=14-2 x 7=-7.

The correct option is (b)

14+8d=70,

=> d=7.

First term is 14-3d=14-2 x 7=-7.

The correct option is (b)

Question 2 |

By adding the same constant to each of 31, 7, â€“1 a geometric progression results. The common ratio is:

$ \displaystyle 13$ | |

$ \displaystyle 2\frac{1}{3}$ | |

$ \displaystyle -12$ | |

$ \displaystyle None\,\,of\,\,these$ |

Question 2 Explanation:

$ \begin{array}{l}(x+31)(x-1)={{(x+7)}^{2}}\\=>{{x}^{2}}+14x+49={{x}^{2}}+30x-31\\=>16x=80\\=>x=5\end{array}$

Question 3 |

The Value of

$ \displaystyle \begin{array}{l}\frac{3}{{{1}^{2}}{{.2}^{2}}}+\frac{5}{{{2}^{2}}{{.3}^{2}}}+\frac{7}{{{3}^{2}}{{.4}^{2}}}+\frac{9}{{{4}^{2}}{{.5}^{2}}}+\frac{11}{{{5}^{2}}{{.6}^{2}}}\\+\frac{13}{{{6}^{2}}{{.7}^{2}}}+\frac{15}{{{7}^{2}}{{.8}^{2}}}+\frac{17}{{{8}^{2}}{{.9}^{2}}}+\frac{19}{{{9}^{2}}{{.10}^{2}}}\,\,is\end{array}$

$ \displaystyle \begin{array}{l}\frac{3}{{{1}^{2}}{{.2}^{2}}}+\frac{5}{{{2}^{2}}{{.3}^{2}}}+\frac{7}{{{3}^{2}}{{.4}^{2}}}+\frac{9}{{{4}^{2}}{{.5}^{2}}}+\frac{11}{{{5}^{2}}{{.6}^{2}}}\\+\frac{13}{{{6}^{2}}{{.7}^{2}}}+\frac{15}{{{7}^{2}}{{.8}^{2}}}+\frac{17}{{{8}^{2}}{{.9}^{2}}}+\frac{19}{{{9}^{2}}{{.10}^{2}}}\,\,is\end{array}$

$ \displaystyle \frac{1}{100}$ | |

$ \displaystyle \frac{99}{100}$ | |

$ \displaystyle \frac{101}{100}$ | |

$ \displaystyle 1$ |

Question 3 Explanation:

$ \begin{array}{l}\frac{3}{{{1}^{2}}{{.2}^{2}}}+\frac{5}{{{2}^{2}}{{.3}^{2}}}+\frac{7}{{{3}^{2}}{{.4}^{2}}}+\frac{9}{{{4}^{2}}{{.5}^{2}}}\\+\frac{11}{{{5}^{2}}{{.6}^{2}}}+\frac{13}{{{6}^{2}}{{.7}^{2}}}+\frac{15}{{{7}^{2}}{{.8}^{2}}}+\frac{17}{{{8}^{2}}{{.9}^{2}}}+\frac{19}{{{9}^{2}}{{.10}^{2}}}\,\\=\frac{4-1}{{{1}^{2}}{{.2}^{2}}}+\frac{9-4}{{{2}^{2}}{{.3}^{2}}}+\frac{16-9}{{{3}^{2}}{{.4}^{2}}}+\frac{25-16}{{{4}^{2}}{{.5}^{2}}}\\+\frac{36-25}{{{5}^{2}}{{.6}^{2}}}+\frac{49-36}{{{6}^{2}}{{.7}^{2}}}+\frac{64-49}{{{7}^{2}}{{.8}^{2}}}\\+\frac{81-64}{{{8}^{2}}{{.9}^{2}}}+\frac{100-81}{{{9}^{2}}{{.10}^{2}}}\\=\frac{1}{{{1}^{2}}}-\,\frac{1}{{{2}^{2}}}+\frac{1}{{{2}^{2}}}-\,\frac{1}{{{3}^{2}}}+\frac{1}{{{3}^{2}}}-\,\frac{1}{{{4}^{2}}}\\+\frac{1}{{{4}^{2}}}-\,\frac{1}{{{5}^{2}}}+\frac{1}{{{5}^{2}}}-\,\frac{1}{{{6}^{2}}}\\+\frac{1}{{{6}^{2}}}-\,\frac{1}{{{7}^{2}}}+\frac{1}{{{7}^{2}}}-\,\frac{1}{{{8}^{2}}}\\+\frac{1}{{{8}^{2}}}-\,\frac{1}{{{9}^{2}}}+\frac{1}{{{9}^{2}}}-\,\frac{1}{{{10}^{2}}}\\=1-\frac{1}{100}\\=\frac{99}{100}\end{array}$

Question 4 |

The value of

$ \displaystyle 1-\frac{1}{20}+\frac{1}{{{20}^{2}}}-\frac{1}{{{20}^{3}}}+.....$ Correct to 5 places f decimal is:

$ \displaystyle 1-\frac{1}{20}+\frac{1}{{{20}^{2}}}-\frac{1}{{{20}^{3}}}+.....$ Correct to 5 places f decimal is:

1.05 | |

0.9 | |

2.9 | |

0.5 |

Question 4 Explanation:

The series is a G.P. and the common ratio is (-1/20) .

Thus the infinite GP sum is (20/21) = 0.9

Thus the correct option is (b)

Thus the infinite GP sum is (20/21) = 0.9

Thus the correct option is (b)

Question 5 |

For all integral values of n,

the largest number that exactly divides each number of the sequence

$ \displaystyle \begin{array}{l}\left( n-1 \right)\,n\,\left( n+1 \right)\,,\,\,\\n\,\left( n+1 \right)\,\left( n+2 \right),\,\,\\\left( n+1 \right)\,\left( n+2 \right)\,\left( n+3 \right)\,.......is\end{array}$

the largest number that exactly divides each number of the sequence

$ \displaystyle \begin{array}{l}\left( n-1 \right)\,n\,\left( n+1 \right)\,,\,\,\\n\,\left( n+1 \right)\,\left( n+2 \right),\,\,\\\left( n+1 \right)\,\left( n+2 \right)\,\left( n+3 \right)\,.......is\end{array}$

12 | |

6 | |

3 | |

2 |

Question 5 Explanation:

Any three consecutive numbers is divisible by 6. Thus the correct option is (b).

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