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I’ve typed the question, in whose context my doubt is, and it’s answer at the end.

Please note that I do not require the solution as I’ve already understood how to find the answer via the given as well as other methods.

My actual question is:

$ 4(p+1)=7q $ and $(p+1)/7=q/4$ are essentially the same equations.

So why don’t we directly do $4(p+1)=7q=k$ , here k is the proportionality constant right?

Why do we get a wrong answer if we use any other form of the equation $(p+1)/7=q/4$ while equating it with k?

For ex:

$4(p+1)=7q$

$4(p+1)/7=q=k$

$p=(7k/4)-1$ and $q=k$

$(7k/4)-1≤102$ and $k≤102$as p,q≤102

$k≤58.86$ and $k≤102$

$=>k≤58$

which is obviously wrong.

Question

Find the number of terms common to the two AP’s: 3,7,11…407 and 2,9,16,..709.

Answer

Let number of terms of two AP’s be m and n respectively.

$ 407=3+(m-1)*4$ and $709=2+(n-1)*7$

$=> m=102$ and $n=102 $

Let pth term of first AP and qth term of second AP be identical.

$3+(p-1)*4=2+(q-1)*7$

$4p-1=7q-5$

$4(p+1)=7q$

$(p+1)/7=q/4=k(say)$

$=> p=7k-1 and q=4k $

As max no. of terms for both AP’s is 102, p,q≤102.

$=>7k-1≤102$ and $4k≤102$

$=>k≤14.71$ and $k≤25.5$

$=> k≤14$ and for each value of k there exists a pair of identical terms. Hence, there are 14 identical terms.

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2 Answers

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Question

Find the number of terms common to the two AP’s $$3,7,11,\dots, 407$$and $$2,9,16,\ldots,709.$$

Answer

Let number of terms of two AP’s be $m$ and $n$ respectively.

$$m=102=n.$$Let the $p$th term of the first AP and the $q$th term of the second AP be identical.$$3+4(p-1)=2+7(q-1)\\4p-1=7q-5$$

Hence, we require the number of integer solutions, with $$1\leq \;p,q\;\leq102,\tag1$$ of the Diophantine equation $$7q-4p=4.\tag{*}$$

By observation^†, $(p,q)=(6,4)$ is a particular solution. Thus, by this theorem, each solution is given by $$(p,q)=(6-7k,4-4k)\quad\text{for some }k\in\mathbb Z.\tag2$$ Think of $k$ as a counter for the solutions.

Solving $(1)$ and $(2)$ gives $$-13.71\leq k\leq0.71.$$ Thus, there are $14$ integer solutions.

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^†Let $n\in\mathbb N$ and $a,b\in\mathbb Z.$ Then, if (and only if) $b$ is a multiple of $\gcd(a,n),$ $$ax+ny=b$$ has an integer solution $(x,y),$ which can be obtained using the Euclidean algorithm.

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The $102$nd term of AP1 is $3+4\cdot101=407$. The $102$nd term of AP2 is $709$. So common terms in these (finite) sequences could have values up to $407$. These common values in the sequences are what your $k$ represents. So values of $k$ could go up as high as $407$.

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