Given sum of first 3 terms of first AP=9
applying formula of sum of AP
9= $\frac{3}{2}*\left ( 2a_{1}+\left ( 2 \right )d_{1} \right )$
9=3* (a1+d1)
a1+d1=3
If second term of an AP is equal to the 9th term of another
a1+d1=a2+8d2
a2+8d2=3
now
sum of 17 terms of second AP
Sn= $\frac{17}{2}*\left ( 2a_{2}+16d_{2} \right )$
Sn=17$\left ( a_{2}+8d_{2} \right )$
Sn=51
Answer option D