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## Algebra Level 2 Test 4

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*Algebra Level 2 Test 4*.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%% Your answers are highlighted below.

Question 1 |

Given that 1 +2+3+4……+10 = 55 , then what will be the sum of 6 + 12 + 18 + 24 …… + 60 ?

A | 300 |

B | 500 |

C | 330 |

D | 306 |

Question 1 Explanation:

We are given the equation 6 + 12 + 18 + 24 …… + 60

We can write this as 6 (1 +2+3+4……+10) = 6 x 55 = 330

Hence the right option is (a)

We can write this as 6 (1 +2+3+4……+10) = 6 x 55 = 330

Hence the right option is (a)

Question 2 |

If 1

^{2}+2^{2}+3^{2}+4^{2}….. +20^{2}= 2870, then what will be the value of 2^{2}+ 4^{2}+ 6^{2}+……………………..+40^{2}?A | 11480 |

B | 11580 |

C | 11680 |

D | 11780 |

Question 2 Explanation:

We are given by the series 1

+2

We need to figure out: 2

This equation can be written as

2

^{2}+2

^{2}+3^{2}+4^{2}….. +20^{2}=2870We need to figure out: 2

^{2}+ 4^{2}+ 6^{2}+……………………..+40^{2}This equation can be written as

2

^{2}x (1^{2}+2^{2}+3^{2}+4^{2}….. +20^{2}) = 4 x 2870 = 11480Question 3 |

The value of 1 – 1/20 + 1/20

^{2}– 1/20^{3 }+ ….. , correct up to 5 place of decimal isA | 1.05 |

B | 0.95238 |

C | 0.95239 |

D | 10.5 |

Question 3 Explanation:

The series which is given is a Geometric Progression series

So we know the sum of an infinite G.P.= when r<1 then sum will be = a/1-r

Where a is the first term and r is the common ratio

So the sum of the series = a/1-r = [1/{1-(-1/20)}]= 20/21 = 0.95238

So we know the sum of an infinite G.P.= when r<1 then sum will be = a/1-r

Where a is the first term and r is the common ratio

So the sum of the series = a/1-r = [1/{1-(-1/20)}]= 20/21 = 0.95238

Question 4 |

If the third term of a G.P. is 3 , then the product of first five terms of G.P. is

A | 81 |

B | 49 |

C | 342 |

D | 243 |

Question 4 Explanation:

Let the G.P. = a/r

Given that the third term is 3

So the series will become

= 3/r

= product of all of these = 243

^{2,}a/r, a, ar, ar^{2}Given that the third term is 3

So the series will become

= 3/r

^{2 }, 3/r , 3 , 3r, 3r^{2}= product of all of these = 243

Question 5 |

If in a pair of fraction, the first fraction is twice the second fraction and the product of the two fraction is 2/25. What will be the value of the fraction?

A | 1/5 |

B | 2/5 |

C | 1/25 |

D | 2/25 |

Question 5 Explanation:

We are told that the first fraction is twice the second fraction

So let the second fraction first = 2p/q

Now the first fraction will be = p/q

On multiplication of two of the fractions we get p/q = 1/5

Therefore first fraction will be = 2 x 1/5 = 2/5

So let the second fraction first = 2p/q

Now the first fraction will be = p/q

On multiplication of two of the fractions we get p/q = 1/5

Therefore first fraction will be = 2 x 1/5 = 2/5

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