# Arithmetic Progression Problem

The sum of 1st six terms of an Arithmetic Progression is 42 the ratio of its 10th term to its 30th term is 1:3. Calculate the first and 13th term of the Arithmetic Progression?

Given that,

• Sum of first 6 terms of the given AP is 42
• a10 : a30 = 1:3

So, Let...

According to the ratio, a10 = 1k =k

a30 = 3k

We know that,

• Sn = n/2(a + l)   {where, Sn= Sum of AP till term n, a = First term of AP, l = last term of AP(also known as an) }
• an = a + (n-1)d   {where, an = Any no. of given AP of nth term, d = Common difference of the consecutive numbers of the AP, n = Term no.}

Now I want to know that how we can 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

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$$a_1=d=2$$

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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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Ok, thanks for the answer –  Alpha Oct 29 '12 at 23:18
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, $a_1 = d= 2$.