# Thread: Arithmetic sequence

1. ## Arithmetic sequence

Pls someone help me:

{Xn },n = 0,1,2,3, is an arithmetic sequence such that the 7th term minus the 4th term is 12 and the sum of the 7th term and 4th terms is -28.Obtain the closed form of the sequence and write down the 10th and 20th term.

Tks

2. Hello, watcher!

There is an arithmetic sequence such that the 7th term minus the 4th term is 12
and the sum of the 7th term and 4th terms is -28.
Obtain the closed form of the sequence and write down the 10th and 20th terms.
You're expected to know this formula . . .

The $n^{th}$ term is: . $a_n \;=\;a + (n-1)d$
. . where $a$ is the first term, and $d$ is the common difference.

The $4^{th}$ term is: . $a_4 \:=\:a + 3d$
The $7^{th}$ term is: . $a_7 \;=\;a + 6d$

Their difference is 12: . $(a+6d)-(a+3d) \:=\:12 \quad\Rightarrow\quad 3d \:=\:12 \quad\Rightarrow\quad\boxed{d = 4}\;\;{\color{red}[1]}$

Their sum is -28: . $(a+6d)+(a+3d)\:=\:\text{-}28 \quad\Rightarrow\quad 2a+9d \:=\:\text{-}28\;\;{\color{red}[2]}$
. . Substitute [1] into [2]: . $2a + 9(4) \:=\:\text{-}28 \quad\Rightarrow\quad\boxed{a \:=\:\text{-}32}$

Hence, the $n^{th}$ term is: . $a_n \;=\;-32 + (n-1)4 \quad\Rightarrow\quad {\color{blue}a_n \:=\:4n-36}$

The $10^{th}$ term is: . $a_{10} \;=\;4(10)-36 \quad\Rightarrow\quad{\color{blue}a_{10}\;=\;4}$

The $20^{th}$ term is: . $a_{20} \;=\;4(20)-36 \quad\Rightarrow\quad{\color{blue}a_{20}\;=\;44}$

3. ## Reply to Soroban

Thanks a lot. But I have a question though and am confused and curious to know (am not so good in math).

Since n starts with zero, why was the formula of Xn = a + (n-1)d used in this problem - and not the closed form of Xn = a + nd (n = 0, 1, 2, ...)?

Thanks again for your help.

Originally Posted by Soroban
Hello, watcher!

You're expected to know this formula . . .

The $n^{th}$ term is: . $a_n \;=\;a + (n-1)d$
. . where $a$ is the first term, and $d$ is the common difference.

The $4^{th}$ term is: . $a_4 \:=\:a + 3d$
The $7^{th}$ term is: . $a_7 \;=\;a + 6d$

Their difference is 12: . $(a+6d)-(a+3d) \:=\:12 \quad\Rightarrow\quad 3d \:=\:12 \quad\Rightarrow\quad\boxed{d = 4}\;\;{\color{red}[1]}$

Their sum is -28: . $(a+6d)+(a+3d)\:=\:\text{-}28 \quad\Rightarrow\quad 2a+9d \:=\:\text{-}28\;\;{\color{red}[2]}$
. . Substitute [1] into [2]: . $2a + 9(4) \:=\:\text{-}28 \quad\Rightarrow\quad\boxed{a \:=\:\text{-}32}$

Hence, the $n^{th}$ term is: . $a_n \;=\;-32 + (n-1)4 \quad\Rightarrow\quad {\color{blue}a_n \:=\:4n-36}$

The $10^{th}$ term is: . $a_{10} \;=\;4(10)-36 \quad\Rightarrow\quad{\color{blue}a_{10}\;=\;4}$

The $20^{th}$ term is: . $a_{20} \;=\;4(20)-36 \quad\Rightarrow\quad{\color{blue}a_{20}\;=\;44}$