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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?

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What have you tried so far? It would be particularly helpful for you to describe your progress on a question like this, so that we can post an answer using the same notation you have already been using. – Brett Frankel Oct 29 '12 at 16:08
up vote 1 down vote accepted

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


$$6a_1+15d=42$$ $$a_1+29d=3(a_1+9d)$$finally we get the system $$2a_1+5d=14$$ $$2a_1-2d=0$$ the solutions are$$a_1=d=2$$

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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.

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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.

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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, $a_1 = d= 2$.

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A short cut and a symmetrical approach: $$a_{10}:a_{30}=1:3 \Rightarrow a_n\propto n \Rightarrow a_n=kn\\ S_6=k\sum_{n=1}^6 n=21k=42 \Rightarrow k=2\\ \therefore a_1=2, a_{13}=26\qquad \blacksquare$$

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