if x1,x2,x3 xn are in H P then prove that x1x2+x2x3+x3x4+ +xn-1xn =( - Maths - Sequences and Series

Question

if x1,x2,x3 xn are in H P then prove that x1x2+x2x3+x3x4+
+xn-1xn =( n-1)x1xn

Ajanta Trivedi · Accepted Answer

given: $x_{1}, x_{2}, x_{3},, x_{n}$ are in HP therefore $\frac{1}{x_{1}}, \frac{1}{x_{2}}, \frac{1}{x_{3}},, \frac{1}{x_{n}}$ are in AP therefore let the common difference of the AP is d therefore $\frac{1}{x_{2}} - \frac{1}{x_{1}} = d \Rightarrow \frac{x_{1} - x_{2}}{x_{1} x_{2}} = d \Rightarrow x_{1} x_{2} = \frac{x_{1} - x_{2}}{d}$ (1) similarly $\frac{1}{x_{3}} - \frac{1}{x_{2}} = d \Rightarrow \frac{x_{2} - x_{3}}{x_{2} x_{3}} = d \Rightarrow x_{2} x_{3} = \frac{x_{2} - x_{3}}{d}$ (2) and $\frac{1}{x_{n}}$ is the nth term of the AP series therefore $\frac{1}{x_{n}} = \frac{1}{x_{1}} + (n - 1) d \frac{1}{x_{n}} - \frac{1}{x_{1}} = (n - 1) d \frac{x_{1} - x_{n}}{x_{n} x_{1}} = (n - 1) d \frac{x_{1} - x_{n}}{d} = (n - 1) x_{1} x_{n}$ (3) now the LHS part of the given equation is: $x_{1} x_{2} + x_{2} x_{3} + x_{3} x_{4} + + x_{n - 1} x_{n} = \frac{x_{1} - x_{2}}{d} + \frac{x_{2} - x_{3}}{d} + \frac{x_{3} - x_{4}}{d} + + \frac{x_{n - 1} - x_{n}}{d} = \frac{1}{d} [x_{1} - x_{2} + x_{2} - x_{3} + x_{3} - x_{4} + + x_{n - 1} - x_{n}] = \frac{1}{d} [x_{1} - x_{n}] = (n - 1) x_{1} x_{n} [f r o m e q (3)]$ = RHS hope this helps you cheers!!

if x1,x2,x3 ....xn are in H.P then prove that x1x2+x2x3+x3x4+....+xn-1xn =( n-1)x1xn