# Arithmetic Progression X.x

• May 5th 2008, 04:26 AM
Tangera
Arithmetic Progression X.x
Thank you for helping! I am soooooooo frustrated with arithmetic and geometric progression because I keep getting stuck at it. sigh.

Q1: The sum to n terms of a particular series is given by Sn = 17n - 3(n^2). Show that the series is an arithmetic progression and state the value of it's common difference.

Q2: The sum of the frist 3 terms and that of the last three terms of an arithmetic progression are 9 and 57 respectively. It is known that the sum of the whole sequence is 121. Find the number of terms in this sequence.

Thank you!
• May 5th 2008, 04:38 AM
mr fantastic
Quote:

Originally Posted by Tangera
Thank you for helping! I am soooooooo frustrated with arithmetic and geometric progression because I keep getting stuck at it. sigh.

Q1: The sum to n terms of a particular series is given by Sn = 17n - 3(n^2). Show that the series is an arithmetic progression and state the value of it's common difference.

Q2: The sum of the frist 3 terms and that of the last three terms of an arithmetic progression are 9 and 57 respectively. It is known that the sum of the whole sequence is 121. Find the number of terms in this sequence.

Thank you!

Q1 From the given series: $\displaystyle s_{n+1} - s_n = 14 - 6n$ .... (1)

But $\displaystyle s_{n+1} - s_n = a_1 + nd$ .... (2)

Compare (1) and (2) ......
• May 5th 2008, 04:52 AM
mr fantastic
Quote:

Originally Posted by Tangera
Q2: The sum of the frist 3 terms and that of the last three terms of an arithmetic progression are 9 and 57 respectively. It is known that the sum of the whole sequence is 121. Find the number of terms in this sequence.

Thank you!

A brutal approach:

"The sum of the frist 3 terms .... of an arithmetic progression are 9":

$\displaystyle a_1 + (a_1 + d) + (a_1 + 2d) = 3a_1 + 3d = 3 \Rightarrow a_1 + d = 1 \Rightarrow {\color{red}a_1 = 1 - d}$ .... (1)

"The sum of ..... the last three terms of an arithmetic progression are ..... 57":

$\displaystyle (a_1 + [n-3]d) + (a_1 + [n-2]d) + (a_1 + [n-1]d) = 3a_1 + 3nd - 6d = 57 \Rightarrow {\color{red}a_1 + nd - 2d = 19}$ .... (2)

"It is known that the sum of the whole sequence is 121":

$\displaystyle 121 = \frac{n(2a_1 + [n-1]d)}{2} \Rightarrow {\color{red}242 = 2na_1 + n(n-1)d}$ .... (3)

Substitute (1) into (2): $\displaystyle d = \frac{18}{n-3}$ .... (4)

Substitute (1) and (4) into (3) and solve for n.
• May 5th 2008, 05:10 AM
Tangera
Thank you very very much for your fast reply! I will continue working on Q1 now...:D
• May 5th 2008, 05:28 AM
Tangera
Quote:

Originally Posted by mr fantastic
A brutal approach:

"The sum of the first 3 terms .... of an arithmetic progression are 9":

$\displaystyle a_1 + (a_1 + d) + (a_1 + 2d) = 3a_1 + 3d = 3 \Rightarrow a_1 + d = 1 \Rightarrow {\color{red}a_1 = 1 - d}$ .... (1)

I was thinking...shouldn't the equation be 3a + 3d = 9??
• May 5th 2008, 05:34 AM
mr fantastic
Quote:

Originally Posted by mr fantastic
A brutal approach:

"The sum of the frist 3 terms .... of an arithmetic progression are 9":

$\displaystyle a_1 + (a_1 + d) + (a_1 + 2d) = 3a_1 + 3d = {\color{blue} 9} \Rightarrow a_1 + d = {\color{blue}3} \Rightarrow {\color{red}a_1 = {\color{blue}3} - d}$ .... (1)

"The sum of ..... the last three terms of an arithmetic progression are ..... 57":

$\displaystyle (a_1 + [n-3]d) + (a_1 + [n-2]d) + (a_1 + [n-1]d) = 3a_1 + 3nd - 6d = 57 \Rightarrow {\color{red}a_1 + nd - 2d = 19}$ .... (2)

"It is known that the sum of the whole sequence is 121":

$\displaystyle 121 = \frac{n(2a_1 + [n-1]d)}{2} \Rightarrow {\color{red}242 = 2na_1 + n(n-1)d}$ .... (3)

Substitute (1) into (2): $\displaystyle d = \frac{{\color{blue}16}}{n-3}$ .... (4)

Substitute (1) and (4) into (3) and solve for n.

I made one stupid mistake that has had a ripple effect. I've made the corrections in blue.
• May 5th 2008, 05:40 AM
mr fantastic
Quote:

Originally Posted by Tangera
I was thinking...shouldn't the equation be 3a + 3d = 9??

Indeed. I was making the corrections while you spotted and posted this. That's what I get for being too quick! I get n = 11 as the answer by the way.
• May 5th 2008, 05:51 AM
Tangera
Okay...me too! yay. Thank you!
• May 5th 2008, 06:09 AM
Soroban
Hello, Tangera!

These are challenging problems . . . Thank you!

Quote:

Q1) The sum to $\displaystyle n$ terms of a particular series is given by: $\displaystyle S_n \:= \:17n - 3n^2$
Show that the series is an A.P. and find its common difference.

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

We must hammer $\displaystyle S_n \:=\:17n - n^2$ into this form

We have: .$\displaystyle S_n \;=\;17n - 3n^2 \;=\;\frac{n}{2}\bigg[34 -6n\bigg]$

Add and subtract 6: .$\displaystyle S_n \;=\;\frac{n}{2}\bigg[34 - 6n \:{\color{blue}+\: 6 - 6}\bigg] \;=\;\frac{n}{2}\bigg[28 - 6(n-1)\bigg]$

And we have: .$\displaystyle S_n \;=\;\frac{n}{2}\bigg[2(14) + (n-1)(\text{-}6)\bigg]$

. . Therefore: .$\displaystyle \boxed{a = 14}\;\;\;\boxed{d = -6}$

The AP is: .$\displaystyle 14,\,8,\,2,\,\text{-}4,\,\text{-}10,\,\text{-}16, \hdots$

Quote:

Q2) The sum of the first 3 terms and that of the last three terms of an A.P.
are 9 and 57, respectively. It is known that the sum of the whole sequence is 121.
Find the number of terms in this sequence.

I hope there is a shorter solution!

The first three terms are: .$\displaystyle a,\;a+d,\;a+2d$
Their sum is 9: .$\displaystyle 3a + 3d \:=\:9\quad\Rightarrow\quad a + d \:=\:3 \quad\Rightarrow\quad a \:=\:3-d\;\;{\color{blue}[1]}$

The last three terms are: .$\displaystyle a + (n-3)d,\;a+(n-2)d,\;a+(n-1)d$
Their sum is 57: .$\displaystyle 3a + 3(n-2)d \:=\:57\quad\Rightarrow\quad a + (n-2)d \:=\:19\;\;{\color{blue}[2]}$

The entire sum is 121: .$\displaystyle \frac{n}{2}[2a + (n-1)d] \:=\:121\quad\Rightarrow\quad n[2a + (n-1)d \:=\:242\;\;{\color{blue}[3]}$

Substitute [1] into [2]: .$\displaystyle (3-d) + (n-2)d \:=\:19\quad\Rightarrow\quad d \:=\:\frac{16}{n-3}\;\;{\color{blue}[4]}$

Substitute [1] into [3]: .$\displaystyle n[2(3-d) + (n-1)d] \:=\:242 \quad\Rightarrow\quad 6n + n(n-3)d \:=\:242\;\;{\color{blue}[5]}$

Substitute [4] into [5]: ,$\displaystyle 6n + n(n-3)\!\cdot\!\frac{16}{n-3} \:=\:242 \quad\Rightarrow\quad 22n \:=\:242$

. . Therefore: . $\displaystyle \boxed{n \:=\:11}$

The AP is: .$\displaystyle 1,3,5,7,9,11,13,15,17,19,21$

Too fast for me, Mr. F . . . Good job!
• May 5th 2008, 06:12 AM
Tangera
Thank you for your help too, Soroban! :)