Sol. Let the A.P. be a 1 , a 2 , ……a n and the G.P. be b 1 , b 2 , ……,b n . We are given that a 1 > 0, b i > 0 for all i, a 1 = b 1 , a 2 = b 2 . We have to show that a n ≤ b n for all n > 2. Let d be common difference of A.P and r be the common ratio of G.P. Then a 2 = a 1 + d and b 2 = a 1 r. Now since all the terms in A.P. are positive we must have d > 0) (In a finite A.P., the assertion may be false).

From a 2 = b 2 , we get, r = 1 + which shows that r > 1

Since d > 0, a 1 > 0.

Now we must show that a 1 + (n –1) d ≤ a 1 r n–1 or, a 1 + (n–1) (a 1 r –a 1 ) ≤ a 1 r n–1

or, 1 + (n –1) (r–1) ≤ r n–1 (Because a 1 > 0)

or, n –1 ≤ (Because r – 1 > 0)

or, n –1 ≤ 1 + r + r 2 +……r n–2

The last inequality we have written is essentially true since R.H.S. contains n –1 terms each of which is greater than unity. ThuS
a n ≤ b n if n > 2.