Given, {a}_{1},{a}_{2},{a}_{3}......&{b}_{1},{b}_{2},{b}_{3}.....are in A.P
Also given, a1=2,a10=3
and a1b1=1=a10b10
So b1=a11=21 and b10=a101=31
Now using nth term formula we get, a10=a1+9da⇒3=2+9da
⇒da=91
Using same formula we get,
b10=b1+9db ⇒31=21+9db ⇒db=−541
Now calculating a4b4=(a1+3da)(b1+3db)
=(2+3×91)(21+3×(54−1))
=37×188=5456=2728