# Thread: Arithmetic Sequences - T5 = 3T2 and S6= 36, find the first three terms.

1. ## Arithmetic Sequences - T5 = 3T2 and S6= 36, find the first three terms.

Yo, some questions.

As in how do u do these. Thanks.

T5 = 3T2 and S6= 36, find the first three terms.

also

Sn = 5n2(sqaured that is) + 3n - Find the first three terms and thus find a formula for Tn

Sweet as Thanks.

2. $\displaystyle a_5=a_2+(5-2)d$ and we know that $\displaystyle a_5=3a_2$ so $\displaystyle 2a_2=3d$. And we were also given: $\displaystyle 36=S_6=\frac {6(a_1+a_6)}2=3((a_2-d)+(a_2+4d))=3(2a_2+3d)=18d\rightarrow d=2$. Using this: $\displaystyle a_2=6$. So the first three terms: $\displaystyle 4, 6, 8$.

For the second problem: $\displaystyle S_n=n(3+5n)=\frac{n(6+10n)}2=\frac{n(6+10(n-1)+10)}2=\frac{n(2\cdot 8+(n-1)\cdot 10)}2$. You can see that $\displaystyle a_1=8$ and $\displaystyle d=10$ therefore $\displaystyle a_n=8+(n-1)\cdot 10$ and the first three terms: $\displaystyle 8, 18, 28$.

3. Hello, Dev!

We are expected to know these formulas:

. . $\displaystyle n^{th}\text{ term: }\;T_n \:=\:a + (n-1)d$

. . $\displaystyle \text{Sum of first }n\text{ terms: }\;S_n \;=\;\frac{n}{2}[2a + (n-1)d]$

. . . . where: .$\displaystyle \begin{Bmatrix}a &=& \text{first term} \\ d &=& \text{common diff.} \\ n &=& \text{no. of terms} \end{Bmatrix}$

$\displaystyle T_5 \:=\: 3\!\cdot\!T_2\:\text{ and }\:S_6\:=\: 36$
Find the first three terms.
2nd term: .$\displaystyle T_2 \:=\:a+d$
5th term: .$\displaystyle T_5 \:=\:a+4d$

$\displaystyle T_5 = 3\!\cdot\!T_2 \quad\Rightarrow\quad a + 4d \:=\:3(a+d) \quad\Rightarrow\quad d \:=\:2a$ .[1]

$\displaystyle S_6 = 36\!:\;\;36 \:=\:\frac{6}{2}[2a + 5d] \:=\:36 \quad\Rightarrow\quad 2x + 5d \:=\:12$ .[2]

Substitute [1] into [2]: .$\displaystyle 2a + 5(2a) \:=\:12 \quad\Rightarrow\quad 12a \:=\:12 \quad\Rightarrow\quad\boxed{ a \:=\:1}$

Substitute into [1]: .$\displaystyle d = 2(1) \quad\Rightarrow\quad\boxed{d \:= \:2}$

The first term is 1, the common difference is 2.

Therefore, the first three terms are: .1, 3, 5.

4. Hello again, Dev!

$\displaystyle S_n \:=\: 5n^2 + 3n$
Find the first three terms and thus find a formula for $\displaystyle T_n$
The sum of the first $\displaystyle n$ terms is: .$\displaystyle S_n \:=\:\frac{n}{2}[2a + (n-1)d]$

The sum of the first one term is: .$\displaystyle a$
The formula says: .$\displaystyle S_1 \:=\:5(1^2) + 3(1) \:=\:8$
. . Hence: .$\displaystyle \boxed{ a\:=\:8}$ .[1]

The sum of the first two terms is: .$\displaystyle a + (a+d) \:=\:2a+d$
The formula says: .$\displaystyle S_2 \:=\:5(2^2) + 3(2) \:=\:26$
. . Hence: .$\displaystyle 2a+d \:=\:26$ .[2]

Substitute [1] into [2]: .$\displaystyle 2(8) + d \:=\:26 \quad\Rightarrow\quad\boxed{ d \:=\:10}$

The first term is: $\displaystyle a = 8$ . . . the common difference is: $\displaystyle d = 10$

$\displaystyle \text{The first three terms are: }\;{\bf{\color{red}8,\:18,\:28}}$

$\displaystyle \text{The }n^{th}\text{ term is: }\;T_n \:=\:8 +10(n-1) \quad\Rightarrow\quad {\bf{\color{red} T_n \:=\:10n -2}}$