The age of the father of two children is twice that of the elder one added to four times that of the younger one. If the geometric mean of the ages of the two children is 4? 3 and their harmonic mean is 6, then what is the father's age?

This is a GP with a = 5 , r = 15/5 = 3, n = 7
S_n = a(rⁿ - 1 )/(r - 1)
? S₁₅ = 5(3⁷ - 1)/(3 - 1)
= 5/2(3⁷ - 1)
= 5/2 (2187 - 1)
= ? 5465

The number of bacteria at any given time forms a GP whose terms are given by 50, 100, 200, ... where
a = 50, r = 100/50 = 2
T_n = ar^{n -1}
? T₁₂ = 50(2)^{12 - 1}
= 50 x (2)¹¹
= 102400
So, the number of bacteria born in 12th hour is 102400.

As, S₁₂ = S₁₈
So, S₁₁ = S₁₉
S₁₀ = S₂₀
S₉ = S₁₉
.........
.........
.........
S₀ = S₃₀
As, S₀ = 0
So, S₃₀ = 0

As, n is odd.
? Sum of even number = [(n - 1)/2 ] [ (n + 1 )/2]
= 280/2 x 282/2 = 19740

Sum of even numbers = n/2[(n/2) + 1)] = (1672/2) x [(1672/2) + 1]
= 699732

The nth term of a GP is ar^{n - 1} ,
5th term = ar^{5 - 1} = ar⁴ = 81
1st term = a = 16
? r⁴ = 81/16
? r = ?81/16 = 3/2
? 4th term = ar^{4 - 1} = ar³ = 16 x 3/2 x 3/2 x 3/2 = 54

Let term = l = ar^{n - 1} a = 5 and l = 20480
r = 20/5 = 4
? 20480 = 5 x (4)^n-1
(4)^n-1 = 20480/5 = 4096 = (4)⁶
n - 1 = 6
? n = 7

(a + b)/2 = 25
a + b = 50
?ab = 7
ab = 49
Hence, A can either be 7 or 49.
So, 49 is the answer.

The second and fourth term of the HP are 1/6 and 1/14 respectively,
Hence, for the corresponding AP, the second term is 6 and fourth term is 14.
Hence a + d = 6 and a + 3d = 14
? 2d = 8
? d = 4 and a = 2

Hence, the 10th term of this AP = a + (10 -1)d
= 2 + (10 - 1)x 4 = 2 + 9 x 4
= 2 + 36 = 38
Hence, for the corresponding HP, the 10th term is 1/38.

T₅ = a + 4d, T₇ = a + 6d
? 5(a + 4d) = 7(a + 6d)
? 5a + 20d = 7a + 42d
? a = -11d
? T₁₂ = a + 11d = - 11d + 11d = 0
So, the twelfth term is 0.