Let the numerically greatest term in the expansion of (3x−16y)15 when x=‌
2
3
and y=‌
3
2
is Tr+1 Tr+1‌=‌15Cr(3(‌
2
3
))15−r(−16(‌
3
2
))r ‌=‌15Cr(2)15−r(−24)r Clearly, the value the expression will be maximum only when the value r is largest possible even integer. So, r=14‌‌[∵0≤r≤15 and r∈Z] ∴15 th term will be the numerically greatest term.