find the sum of n AM's between two number is n times the single AM between them
Answers
Answer:
Sol:
Let n arthematic means be A1,A2,A3................An between 5 and 35 then
5, A1,A2,A3................An,35 are in A.P And common difference d.
a = 5 and
∴ tn+2 = 5 + ( n + 2 - 1)d = 35
⇒ d = ( 35 - 5) / ( n +1)
⇒ d = 30 / ( n +1).
A1 = t2 = 5 + d = 5 + 30 / ( n +1)
⇒ A1 = t2 = (5n + 35) / ( n +1).
An = tn+1 = 5 + nd = 5 + 30n / ( n +1)
⇒ An = tn+1 = (5 + 35n) / ( n +1).
Given A1 : An = 1: 4
(5n + 35) / ( n +1) : (5 + 35n) / ( n +1) = 1: 4
⇒ (5n + 35) : (5 + 35n) = 1: 4
⇒ (5n + 35) / (5 + 35n) = 1 / 4
⇒ 4(5n + 35) = (5 + 35n)
⇒ 20n + 140 = 5 + 35n
15n = 135
n = 135 / 15.
∴ n = 9.
2)
Let two numbers are a,b and Let A1,A2,A3................An n A.Ms between them.
a, A1,A2,A3................An,b are in A.P.
Let d be the common difference of A.P.
∴ tn+2 = a + ( n + 2 - 1)d = b
⇒ d = ( b - a) / ( n +1).
A1 = t2 = a + d
An = tn+1 = a + nd.
Sum of n A.M’s = A1 + A2 + A3................+An = ( n /2) [A1 + An ]
Sum of n A.M’s = ( n /2) [a + d + a + nd ]
Sum of n A.M’s = ( n /2) [2a + (n + 1)d ].
Sum of n A.M’s = ( n /2) [2a + (n + 1) ( b - a) / ( n +1) ].
Sum of n A.M’s = ( n /2) [2a + ( ( b - a)].
Sum of n A.M’s = ( n /2) [a + b].
Sum of n A.M’s = n A.
Where A is single Arithmetic mean between a and b.
Thus Sum of n A.M’s between a and b is equal to n times Arithmetic mean between a and b.
Step-by-step explanation: