Given that, AP(a ; d) denote the set of all the terms of an infinite arithmetic progression with first term 'a' and common difference d > 0.

Now, let m^th term of first progression

$$AP(1;3) = 1 + (m - 1)3 = 3m - 2$$ .... (i)

and n^th term of progression

$$AP(2;5) = 2 + (n - 1)5 = 5m - 3$$ .... (ii)

and r^th term of third progression

$$AP(3;7) = 3 + (r - 1)7 = 7m - 4$$ .... (iii) are equal.

Then, $$3m - 2 = 5n - 3 = 7r - 4$$

Now, for $$AP(1;3) \cap AP(2;5) \cap AP(3;7)$$,

the common terms of first and second progressions, $$m = {{5n - 1} \over 3}$$

$$ \Rightarrow $$ n = 2, 5, 11, ...

and the common terms of second and the third progressions,

$$r = {{5n + 1} \over 7}$$ $$ \Rightarrow $$ n = 4, 11, ....

Now, the first common term of first, second and third progressions (when n = 11), so

a = 2 + (11 - 1)5 = 52

and d = LCM (3, 5, 7) = 105

So, $$AP(1;3) \cap AP(2;5) \cap AP(3;7)$$ = AP(52; 105)

So, a = 52 and d = 105

$$ \Rightarrow $$ a + d = 157.00