Arithmetic Progression

Let d be the common difference of the AP. We can expand $A_1+A_4+A_7+\dots+A_{16} = 147$ as $A_1+(A_1+3d)+(A_1+6d)+\dots+(A_1+15d)=157$ . This simplifies to $6A_1+45d=147$ .

Similarly, we can express $A_1+A_6+A_{11}+A_{16}$ as $A_1+(A_1+5d)+(A_1+10d)+(A_1+15d)$ , which simplifies to $4A_1+30d$ .

We note that $\frac{2}{3}(6A_1+45d)=4A_1+30d$ , and since $6A_1+45d=147$ ,

$A_1+A_6+A_{11}+A_{16}=4A_1+30d=\frac{2(147)}{3}=98$ .

its too easy to be rated as level 3