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?