Arithmetic series..... other problems..

Nov 2009
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An arithmetic series has first term a and common difference d.
The sum of the first 31 terms of the series is 310.


a) Show that a + 15d = 10

> S31 = 31/2 ( 2a + (31-1) d)
> 31( a + 15d ) = 310
> a + 15d = 310/31; a + 15d = 10
(Rock)

b)Given also that the 21st term is twise the 16th term, find the value of d.

???

c)the nth term of the series is un. Given that
K
Sum un = 0, find the value of K.
n = 1

???
 
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CaptainBlack

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An arithmetic series has first term a and common difference d.
The sum of the first 31 terms of the series is 310.


a) Show that a + 15d = 10

> S31 = 31/2 ( 2a + (31-1) d)
> 31( a + 15d ) = 310
> a + 15d = 310/31; a + 15d = 10
(Rock)

b)Given also that the 21st term is twise the 16th term, find the value of d.
\(\displaystyle 2(a+20d)=a+15d\)

so

\(\displaystyle a+25d=0\)

and you already have:

\(\displaystyle a+15d=10\)

So solve these simultaneous equations.

CB
 

CaptainBlack

MHF Hall of Fame
Nov 2005
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5,271
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c)the nth term of the series is un. Given that
K
Sum un = 0, find the value of K.
n = 1

???
From the previous part you will now know \(\displaystyle a\) and \(\displaystyle b\), so plug them into the formula for the sum, which gives a equation in K to solve:

\(\displaystyle \sum_{n=1}^K u_n=\frac{K}{2}(2a+(K-1)d)=0\)

CB