Arithmetic sequence

**thereddemon** · March 16th 2009, 12:46 PM

The first term of an arithmetic sequence is 30 and the common differenc eis -1.5.
a) Find the value of the 25th term.

The rth term of the sequence is 0.
b)Find the value of r.

The sum of the first n terms of the sequence is Sn.

c) Find the largest postitive value of Sn.

i worked out a and b but got stuck with c. Any help!

**e^(i*pi)** · March 16th 2009, 01:20 PM

Originally Posted by thereddemon

The first term of an arithmetic sequence is 30 and the common differenc eis -1.5.
a) Find the value of the 25th term.

The rth term of the sequence is 0.
b)Find the value of r.

The sum of the first n terms of the sequence is Sn.

c) Find the largest postitive value of Sn.

i worked out a and b but got stuck with c. Any help!

Part c will be the sum of terms from the first term to your answer to part b which will be the sum of all the positive terms hence the most positive

**Soroban** · March 17th 2009, 09:29 AM

Hello, thereddemon!

Another approach . . .

The first term of an arithmetic sequence is $a = 30$
. . and the common difference is $d = \text{-}1.5$

The sum of the first $n$ terms of the sequence is $S_n.$

c) Find the largest postitive value of $S_n.$

Formula: . $S_n \;=\;\frac{n}{2}\bigg[2a + (n-1)d\bigg]$

We have: . $a = 30,\;d = \text{-}1.5$

. . $S \:=\:\frac{n}{2}\bigg[2(30) + n(-1.5)\bigg] \quad\Rightarrow\quad S \:=\:30n - 0.75n^2$

This is a down-opening parabola; its maximum is at its vertex.

. . The vertex is at: . $n \:=\:\frac{\text{-}b}{2a} \:=\:\frac{\text{-}30}{2(\text{-}0.75)} \:=\:20$

Therefore, maximim $S$ is: . $S \:=\:30(20)+0.75(20^2) \;=\;300$

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