Question: > 
> d>0. If $AP(1,3)\cap AP(2,5)\cap AP(3,7)=AP(a,d)$, then the value of a+d is
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> 
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Let the common term of the APs be m, n and r.

For first and second AP
$$1+(m-1)3=2+(n-1)5$$
$$m=\frac{5n-1}{3}$$
And for second and third AP
$$2+(n-1)5=3+(r-1)7$$
$$r=\frac{5n+1}{7}$$
I found the relationships between them , but I couldn’t solve further. Please help.

Answer: 
You showed that $m=\frac{5n-1}{3}$. But we also know that $m$ and $n$ are positive integers. The smallest possible solution to that is $m=3,\ n=2$. The third term of $AP(1,3)$ is $7$, as is the second term of $AP(2,5)$. The next solution is $m=8,\ n=5$, which corresponds to $22$. Carrying on like this, it shouldn't be hard to convince yourself that $$AP(1,3)\cap AP(2,5)=AP(7,15)$$.

If you can carry on this way to find $AP(7,15)\cap AP(3,7)$, you're all done.
$$3m=5n-1\iff3(m+2)=5(n+1)$$

$\implies\dfrac{5(n+1)}3=m+2$ which is an integer

$\implies3|5(n+1)\iff3|(n+1)$

We can write $n+1=3p\iff n=3p-1$ where $p$ is any integer

$$3+7(r-1)=2+5(n-1)=2+5(3p-2)$$

$$\iff7r+4=15p-8\iff7r=15p-2(7\cdot3-15)$$

$$\iff\dfrac{15(p+2)}7=r+6$$ which is an integer

$\implies7|15(p+2)\iff7|(p+2)$

We can set $p+2=7q$ where $q$ is any integer

Now we can find $n,$ hence $m,r$ in terms of $q$