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December 20, 2014

December 20, 2014

Posted by **Anonymous** on Thursday, April 17, 2008 at 12:26pm.

2) Find the least number of terms of the AP, 1+3+5...that are required to make a sum exceeding 4000.

(?a=1,d=2)

Any help at all would be great

- Maths -
**Reiny**, Thursday, April 17, 2008 at 2:18pm1)

remember that t(n) = a+(n-1)d

t(8) = a+7d

t(4) = a+3d

it said: the 8th term is twice the 4th term ---> a+7d = 2(a+3d), simplify this

then it said: the 20th term is 40 ---> a+19d = 40

Solve these two equations.

2) YOu should know that

S(n) = n/2[2a + (n-1)d]

so you want n/2[2 + (n-1)(2)] > 4000

solve this

- Maths -
**ve**, Sunday, August 22, 2010 at 12:43pm2n

- Maths -
**Anonymous**, Wednesday, February 1, 2012 at 6:00amTh 4th term 0f an a p is 13 and the 10th term is 29 find the 16th term

- physics -
**metric and non-metric system**, Monday, January 21, 2013 at 10:19ammetric countries are : The United Kingdom,Canada,New Zealand,Republic of Ireland.

non-metre countries are: The United States,Burma(Myanmar),Liberia.

both metric and non-metric countries are:

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