Question

if x1,x2,x3.....xn are in ap the what is the value of 1/x1x2+1/x2x3+1/x3x4........+1/xn-1xn

Answer 1

Answer:

Step-by-step explanation:

=> Common difference of AP:

d = x2 - x1 = x3 - x2 = x4 - x3 = ... = xn - x(n-1)

xn = x1 + (n-1)d

xn - x1 = (n-1) d

∴ d = (xn - x1) / (n-1)

=> Value = 1/x1x2 + 1/x2x3 + 1/x3x4 + ........ + 1/x(n-1)xn

= (1/d)[d/x1x2 + d/x2x3 + d/x3x4 + ... + d /x(n-1)xn]

= [(n-1)/(xn-x1)][(x2-x1)/x1x2 + (x3-x2)/x2x3 + (x4-x3)/x3x4 + ... + (xn - x(n-1)/x(n-1) xn]

= [(n-1)/(xn-x1)] [1/x1 - 1/x2 + 1/x2 - 1/x3 + 1/x3 - 1/x4 + ... + 1/x(n-1) - 1/xn]

= [(n-1)/(xn-x1)] [1/x1 -1/xn]

= [[(n-1) / (xn-x1)] *  [(xn - x1) / x1xn]

= (n-1) / x1xn

Thus, the value of 1/ x1x2 + 1/x2x3 + 1/x3x4 +...+ 1/xn-1 xn is '(n-1) / x1xn'.