Let a1, a2,…,a10 be in A.P and h1, h2.....h10 be in H.P. If a1=h1=2 and
a10=h10=3, then a4h7 is
2
3
5
6
Let d be the common difference of the A.P…., then
3=a10=2+9d ⇒ d=1/9∴ a1=2+3d=7/3
Next, let D be the common difference of the A.P.
1h1,1h2,…,1h10 Then
13=1h10=12+9D⇒D=−154∴ 1h7=1h1+6D=718⇒h7=187
Hence, a4h7=(7/3)(18/7)=6.