Question

Let $A.P.(a;d)$ denote the set of all the terms of an infinite arithmetic progression with first term $a$ and common difference $d>0$. If $A.P(1;3)\cap A.P(2;5)\cap A.P.(3;7)=A.P(a;d)$, then $a+d$ equals

Answer 1

Let the $n^{th}$ term of first A.P. is equal to $m^{th}$ term of second A.P. and is equal to $k^{th}$ term of third A.P., so
$1+(n-1)3=2+(m-1)5=3+(k-1)7\Rightarrow 3n-2=5m-3=7k-4\Rightarrow m=\frac{3n+1}{5},k=\frac{3n+2}{7}$

As $m,n,k$ are integers, so
For $m$ to be an integer $3n+1$ should be divisible by $5$, so
$n=3,8,13,18,\dots$
For $k$ to be an integer $3n+2$ should be divisible by $7$, so
$n=4,11,18,25,\dots$
So, the least common value of $n=18$

Now, the first term of required A.P
$a=1+(18-1)3=52$
Common Difference of required A.P
$d=\text{L.C.M}(3,5,7)=105$

$\therefore a+d=52+105=157$