# Thread: Arithmetic Progression X.x

1. ## 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!

2. 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) ......

3. 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.

4. Thank you very very much for your fast reply! I will continue working on Q1 now...

5. 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??

6. 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.

7. 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.

8. Okay...me too! yay. Thank you!

9. Hello, Tangera!

These are challenging problems . . . Thank you!

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

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!

10. Thank you for your help too, Soroban!