Let rthand (r + 1)th be two consecutive terms in the expansion of (3 – 5x)15
Tr + 1 ≥ Tr
15Cr 315 – r (| – 5x|)r ≥ 15Cr – 1 315 – (r – 1) (|– 5x|)r – 1
16 – r ≥ 3r
4r ≤ 16 r ≤ 4
Explanation:
For r ≤ 4, Tr + 1 ≥ Tr
⇒ T2 > T1
T3 > T2
T4 > T3
T5 = T4
For r > 5, Tr + 1 < Tr
T6 < T5
T7 < T6
and so on Hence, T4 and T5 are numerically greatest terms and both are equal.
