if a1,a2,a3.....an be an a.p of non zero terms
prove that:1/a1a2+1/a2a3+....+1/an-1an =n-1/a1an
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Let d denote the common difference of successive terms in the arithmetic progression. So a_2 = a_1 + d, a_3 = a_2 + d, and so on.
We can assume that d is nonzero since an arithmetic progression with d = 0 is a constant sequence, and the desired result is obvious in that case (if a_j = c for all j, then the sum 1/(a_1 a_2) + ... + 1/(a_{n-1} a_n) is the sum of n-1 copies of 1/c^2, which is (n - 1)/c^2, which i s (n - 1)/(a_1 a_n)). Now we recall the algebraic identity
1/(x(x+d)) = (1/d) (1/x - 1/(x + d)).
You can verify this identity by simply adding the fractions on the right hand side and seeing that you get the left hand side, or by applying the method of "partial fraction decomposition" to the left hand side. Since a_{j+1} = a_j + d for any j between 1 and n-1, the above identity with x = a_j implies that
1/(a_j a_{j+1}) = (1/d) (1/a_j - 1/a_{j+1}), 1 <= j < n.
If you sum the left hand side of this from j = 1 to j = n-1, it is the same as summing the right hand side from j = 1 to n - 1, which is
(1/d) ((1/a_1 - 1/a_2) + (1/a_2 - 1/a_3) + ... + (1/a_{n-1} - 1/a_n)).
All terms in the sum cancel except the very first, 1/a_1, and the very last, -1/a_n, giving us the result
(1/d)(1/a_1 - 1/a_n) = (1/d) (a_n - a_1)/(a_1 a_n).
Since the sequence is an arithmetic progression, we have a_n = a_1 + (n-1) d, so the right hand side of the above simplifies to (n - 1)/(a_1 a_n).
We can assume that d is nonzero since an arithmetic progression with d = 0 is a constant sequence, and the desired result is obvious in that case (if a_j = c for all j, then the sum 1/(a_1 a_2) + ... + 1/(a_{n-1} a_n) is the sum of n-1 copies of 1/c^2, which is (n - 1)/c^2, which i s (n - 1)/(a_1 a_n)). Now we recall the algebraic identity
1/(x(x+d)) = (1/d) (1/x - 1/(x + d)).
You can verify this identity by simply adding the fractions on the right hand side and seeing that you get the left hand side, or by applying the method of "partial fraction decomposition" to the left hand side. Since a_{j+1} = a_j + d for any j between 1 and n-1, the above identity with x = a_j implies that
1/(a_j a_{j+1}) = (1/d) (1/a_j - 1/a_{j+1}), 1 <= j < n.
If you sum the left hand side of this from j = 1 to j = n-1, it is the same as summing the right hand side from j = 1 to n - 1, which is
(1/d) ((1/a_1 - 1/a_2) + (1/a_2 - 1/a_3) + ... + (1/a_{n-1} - 1/a_n)).
All terms in the sum cancel except the very first, 1/a_1, and the very last, -1/a_n, giving us the result
(1/d)(1/a_1 - 1/a_n) = (1/d) (a_n - a_1)/(a_1 a_n).
Since the sequence is an arithmetic progression, we have a_n = a_1 + (n-1) d, so the right hand side of the above simplifies to (n - 1)/(a_1 a_n).
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