Test: Arithmetic Mean - Commerce MCQ

If a, b, c are in A.P., then b is the arithmetic mean of a and c.
a b and c are in AP.
∴ b - a = c - b
⇒ 2b = a + c          ........ (1)
The arithmetic mean of a and c
= (a+c)/2 
Using equation (1), we get
= 2b/2
= b 
∴ The arithmetic mean of a and c = b
Thus, the arithmetic mean of a and c is b.

a = 4, l = 31, n = 8
(l-a)/(n+1) = (31-4)/(8+1)
= 27/9 
= 3

(aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) is mean between a & b
Mean of a & b =  ( a + b)/2
=> (aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) = ( a + b)/2
=> 2aⁿ⁺¹ + 2bⁿ⁺¹  = aⁿ⁺¹  + bⁿ⁺¹  + baⁿ + bⁿa
=> aⁿ⁺¹ + bⁿ⁺¹ = baⁿ + bⁿa
=> aⁿ(a - b) = bⁿ(a - b)
=> aⁿ = bⁿ
=> (a/b)ⁿ = 1
=> n = 0    or a  = b

Let the digits at ones, tens and hundreds place be (a−d)a and (a+d) respectively. The, the number is
(a+d)×100+a×10+(a−d) = 111a+99d
The number obtained by reversing the digits is
(a−d)×+a×10+(a+d) = 111a−99d
It is given that the sum of the digits is 21.
(a−d)+a+(a+d) = 21                        ...(i)
Also it is given that the number obtained by reversing the digits is 594 less than the original number.
∴111a−99d = 111a+99d−396          ...(ii)
⟹ 3a = 21 and 198d = 396
⟹ a = 7 and d = -2
Original number = (a−d)×+a×10+(a+d)
= 100(9) + 10(7) + 5
= 975

d = 100
S_n = 21700
n = 14
S_n = n/2[2a + (n-1)d]
21700 = 14/2[2a + (14-1)100]
= 21700 = 7[2a + 1300]
= 3100 = 2a + 1300
⇒ 3100 - 1300 = 2a
⇒ 1800 = 2a
⇒ a = 900

2 and 57 have 10 terms between them so including them there would be 12 terms
a_n = 57, a = 2, n = 12
a_n = a + (n-1)d
=> 57 = 2 + (12 - 1)d
=> 55 = 11d
d = 5
T10 = a + 10d
=> 2 + 10(5) 
= 52

 6, x, -12
x = (6-12)/2
x = -6/2
x = -3

A₁,A₂,......, An are inserted between a and b then the series will become 
a, A₁,A₂,A₃,......, An,b. Now a becomes the first term, A₁ will be second, A₂ will become third term
An will become A(n+1)th term 
therefore A(n-1) will become nth term.