Let 6 Arithmetic means A1, A2, A3, A4, A5, As are inserted between two consecutive natural number
and b (a > b). If A12 - A22 + A32 - A1+ As2 - As? is equal to prime number then 'b' is equal to
(A) 1
(B) 2
C3
(D) 4
Answers
Let 6 Arithmetic means A1, A2, A3, A4, A5, As are inserted between two consecutive natural number
and b (a > b). If A12 - A22 + A32 - A1+ As2 - As? is equal to prime number then 'b' is equal to
Given,
6 Arithmetic means A1, A2, A3, A4, A5, A6
If N AMs are inserted between two numbers a and b, then we have common difference,
d = ( b - a ) / (N + 1)
as N = 6, we have,
d = ( b - a ) / (6 + 1)
d = ( b - a ) / 7
Now consider,
A1² - A2² + A3² - A4²+ A5² - A6²
= (A1 + A2) (A1 - A2) + (A3 + A4) (A3 - A4) + (A5 + A6) (A5 - A6)
= (A1 + A2) (-d) + (A3 + A4) (-d) + (A5 + A6) (-d)
= (-d) (A1 + A2 + A3 + A4 + A5 + A6)
as we know that,
A1 = a + d
A2 = a + 2d
A3 = a + 3d
A4 = a + 4d
A5 = a + 5d
A6 = a + 6d
= (-d) (a + d + a + 2d + a + 3d + a + 4d + a + 5d + a + 6d)
= (-d) (6a + 21d)
= - ( b - a ) / 7 [ 6a + 21 ( b - a ) / 7 ]
= (a - b)/7 (21a + 21b)/7
= (a - b)/7 × 3(a + b)
= 3/7 (a - b) (a + b)
= 3/7 (a² - b²)
Given 3/7 (a² - b²) is a prime number
This is possible only when, (a² - b²) = 7
Given a and b are consecutive numbers and a > b. Therefore, a = b + 1
using the above equation, we get,
(a² - b²) = 7
(b+1)² - b² = 7
b² + 1 + 2b - b² = 7
1 + 2b = 7
2b = 6
b = 3
Therefore the values of a and b are a = b + 1 = 3 + 1 = 4 and b = 3
∴ a = 4, b = 3