JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Sequences and Series

We need to find the common terms of the three APs. The general terms are:

• AP(1; 3): $1+3a$
• AP(2; 5): $2+5b$
• AP(3; 7): $3+7c$

Equating: $1+3a=2+5b=3+7c$

From $1+3a=2+5b$: $3a=1+5b$, so 3 divides $1+5b$, which means 3 divides $1+2b$.

From $2+5b=3+7c$: $5b=1+7c$, so 5 divides $1+7c$, which means 5 divides $1+2c$.

The smallest value satisfying $1+2c\equiv 0(\mathrm{mod}5)$ is $c=2$, giving $1+2c=5$. But we need $1+2c=15$ for $c=7$ to get the first common term.

When $c=7$: $3+7(7)=52$

The common difference is $\text{lcm}(3,5,7)=105$.

Therefore: $a+d=52+105=157$