Arithmetic Progression Problem The sum of 1st six terms of an Arithmetic Progression is 42, while the ratio of its 10th term to its $30$th term is $1:3$. 
Calculate the first and the $13$th term of this Arithmetic Progression?
What I'd done yet,
Given that,


*

*Sum of first $6$ terms of the given AP is $42$

*$a_{10}$ : $a_{30}$ = $1:3$


So, Let...
According to the ratio, $a_{10} = 1k =k$
$a_{30} = 3k$
We know that,


*

*$S_{n} = n/2(a + l)$    {where, $S_n$= Sum of AP till term $n$, $a$ = First term of AP, $l$ = last term of AP(also known as $a_{n}$) }

*$a_{n} = a + (n-1)d$   {where, $a_{n}$ = Any no. of given AP of $n_{th}$ term, $d$ = Common difference of the consecutive numbers of the AP, $n$ = Term no.}


Now I want to know that how can I equate it?
 A: In an arithmetic progression $a,a+d,a+2d,...$, the $n$th term is $a+(n-1)d$ and the sum to $n$ terms is $\frac{n}{2}(2a+(n-1)d)$.
If the ratio of the tenth term to the thirtieth term is $\frac{1}{3}$, then $3(a+9d)=a+29d$. If the sum of the first six terms is $42$, then $3(2a+5d)=42)$.
Solve for $a$ and $d$ and then find the first and thirteenth term.
A: In airthmatic progression the terms are a,a+d,a+2d,a+3d,.......a+(n-1)d where 6th term is a+5d
and sum of first six terms is equal to 6a+10d.
and in question it is given that (a+9d)/(a+29d) = 1/3
by these two equations we can easily find out a and d and then you can calculate first and 13th term of an airthmatic prograssion.
A: General therm of AP is $a_n=a_1+(n-1)d$ and sum of first n-therms is$$S_n=\frac{n}{2}(a_1+a_n)=\frac{n}{2}(2a_1+(n-1)d)$$
from conditions we have
$$a_1+a_2+...+a_6=\frac{6}{2}(2a_1+5d)=42$$and$$a_{10}:a_{30}=1:3$$or
$$6a_1+15d=42$$
$$a_1+29d=3(a_1+9d)$$finally we get the system
$$2a_1+5d=14$$
$$2a_1-2d=0$$
the solutions are$$a_1=d=2$$
A: Since $a_{10} : a_{30} = 1 : 3$, we have $\dfrac{a_{10}}{a_{30}-a_{10}}=\dfrac{1}{3-1}$ which implies $a_{10}=\frac{20}{2}d = 10d$ since $a_{30}-a_{10}=(30-10)d$. Therefore $a_1=d$, and so $42 = S_6 = 6d + \frac{6(6-1)}{2} d = 21d$. So, $a_1 = d= 2$.
A: A short cut and a symmetrical approach:
$$a_{10}:a_{30}=1:3 \Rightarrow a_n\propto n \Rightarrow a_n=kn\\
S_6=k\sum_{n=1}^6 n=21k=42 \Rightarrow k=2\\
\therefore a_1=2, a_{13}=26\qquad \blacksquare$$
