Let 3,7,11,15, , 403 and 2,5,8,11, , 404 be two arithmetic progressions.

Question

Let $3,7,11,15, \ldots, 403$ and $2,5,8,11, \ldots, 404$ be two arithmetic progressions. Then the sum, of the common terms in them, is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 6699

Accepted Answer

To find the common terms in the two given arithmetic progressions (AP), we need to first identify the common difference for each sequence and then find the sequence that represents their overlap by employing the concept of least common multiple (LCM).

The first AP is:

$3,7,11,15,\ldots ,403$

The common difference (${d}_{1}$) for the first AP can be calculated by subtracting the first term from the second term:

${d}_{1}=7-3=4$

The second AP is:

$2,5,8,11,\ldots ,404$

The common difference (${d}_{2}$) for the second AP is:

${d}_{2}=5-2=3$

To find the terms common to both sequences, we need to find a term that appears in both sequences. Any common term must be of the form $3+4k$ and $2+3l$ for some integers $k$ and $l$. We want to find when these two forms will give us the same number, so we set them equal to each other:

$3+4k=2+3l$

Rearranging the terms gives us:

$4k-3l=2-3$

This simplifies to:

$4k-3l=-1.......(1)$

The solutions to equation $(1)$ will give us the common terms. Notice this is a Diophantine equation (A Diophantine equation is a polynomial equation, usually with two or more variables,) and has an infinite number of solutions. Let's find one such solution. We can see that:

$k=1\, \, 4(1)-3l=-1\, \implies \, 4-3l=-1\, \implies \, 3l=5\, \implies \, l=1\frac{2}{3}$

This is not an integer solution for $l$, so $k=1$ does not work. Trying $k=2$ gives:

$4(2)-3l=-1\, \implies \, 8-3l=-1\, \implies \, 3l=9\, \implies \, l=3$

Now we've found integers $k=2$ and $l=3$ that satisfy the equation. The corresponding term in both sequences would be:

$3+4(2)=3+8=11\, \, 2+3(3)=2+9=11$

Since $11$ is a common term, we can assert that every common term in both APs will be of the form $11+m(4	imes 3)$, where $m$ is a non-negative integer, and $4	imes 3=12$ is the LCM of the common differences of the two APs. Thus, the general form for the common terms would be:

$11+12m$

Now we are to find all terms that are common up to $403$ in the first sequence and up to $404$ in the second sequence. Because the first sequence doesn't exceed $403$, we'll use this as our limit:

$11+12m\le 403$

To find the largest possible integer value for $m$, we solve the inequality:

$12m\le 403-11$

$12m\le 392$

$m\le 32\frac{2}{3}$

Since $m$ has to be an integer, the largest possible value for $m$ is $32$. Therefore, the common terms are generated by $m=0,1,2,\ldots ,32$. There are $32+1=33$ terms in total.

We will now sum these up. The sum of an AP is given by the formula:

$S=\frac{n}{2}({a}_{1}+{a}_{n})$

Where $S$ is the sum, $n$ is the number of terms, ${a}_{1}$ is the first term, and ${a}_{n}$ is the last term. Using the formula:

$S=\frac{33}{2}(11+(11+12	imes 32))$

$S=\frac{33}{2}(11+11+384)$

$S=\frac{33}{2}(406)$

$S=33	imes 203$

$S=6699$

Therefore, the sum of the common terms in the two arithmetic progressions is 6699.

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