Question

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

Answer 1

The value of $a+d$ is $157$

Explanation for the correct answer:

According to given data

First series $AP(1;3)$ is $\left\{1,4,7,10,13,..\right\}$

Second series $AP(2;5)$ is $\left\{2,7,12,17,...\right\}$

Third series $AP(3;7)$ is $\left\{3,10,17,24,..\right\}$

Finding value of $a+d$

Observing the least number in the third series which leaves remainder $1$ when it is divided by $3$ and also leaves remainder $2$ when it is divided by $5$.

Therefore, we have $52$ such number.

So, considering $a=52$

And $d$ is L.C. M. of $(3,5,7)=105$

Thus,

$$\begin{gathered} a+d=52+105 \\ =157. \end{gathered}$$

Hence, the required value of $a+d=157.$