$${1 \over {{a_2}}} - {1 \over {{a_1}}} = {1 \over {{a_3}}} - {1 \over {{a_2}}} = .........$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over {{a_n}}} - {1 \over {{a_{n - 1}}}} = d$$ (say)

Then $${a_1}{a_2} = {{{a_1} - a{}_2} \over d},\,{a_2}{a_3} = {{{a_2} - {a_3}} \over d},$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......,\,{a_{n - 1}}{a_n} = {{{a_{n - 1}} - {a_n}} \over d}$$

$$\therefore$$ $${a_1}a{}_2 + {a_2}{a_3} + ......... + {a_{n - 1}}{a_n}$$

$$ = {{{a_1} - {a_2}} \over d} + {{{a_2} - {a_3}} \over d} + ......$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {{{a_{n - 1}} - {a_n}} \over d}$$

$$ = {1 \over a}\left[ {{a_1}} \right. - {a_2} + {a_2} - {a_3} + .......$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. { + {a_{n - 1}} - an} \right] = {{{a_1} - {a_n}} \over d}$$

Also, $${1 \over {{a_n}}} = {1 \over {{a_1}}} + \left( {n - 1} \right)d$$

$$ \Rightarrow {{{a_1} - {a_n}} \over {{a_1}{a_n}}} = \left( {n - 1} \right)d$$

$$ \Rightarrow {{{a_1} - {a_n}} \over d} = \left( {n - 1} \right){a_1}{a_n}$$

Which is the required result.

$${{{e^x} + {e^{ - x}}} \over 2}$$

$$ = 1 + {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}} + {{{x^6}} \over {6!}}.........$$

Putting $$x = {1 \over 2}$$ we get

$$1 + {1 \over {4.2!}} + {1 \over {16.4!}} + {1 \over {64.6!}} + .......$$

$$\infty = {{{e^{{1 \over 2}}} + {e^{{{ - 1} \over 2}}}} \over 2} = {{\sqrt e + {1 \over {\sqrt e }}} \over 2}$$

$$ = {{e + 1} \over {2\sqrt e }}$$

$$x = \sum\limits_{n = 0}^\infty {{a^n}} = {1 \over {1 - a}}\,\,\,\,\,\,\,\,\,\,a = 1 - {1 \over x}$$

$$y = \sum\limits_{n = 0}^\infty {{b^n}} = {1 \over {1 - b}}\,\,\,\,\,\,\,\,\,\,b = 1 - {1 \over y}$$

$$z = \sum\limits_{n = 0}^\infty {{c^n}} = {1 \over {1 - c}}\,\,\,\,\,\,\,\,\,\,c = 1 - {1 \over z}$$

$$a,b,c$$ are in $$A.P.$$ OR $$2b = a + c$$

$$2\left( {1 - {1 \over y}} \right) = 1 - {1 \over x} + 1 - {1 \over y}$$

$${2 \over y} = {1 \over x} + {1 \over z} \Rightarrow x,y,z$$ are in $$H.P.$$

We know that

$$e = 1 + {1 \over {1!}} + {1 \over {2!}} + {1 \over {3!}} + ..........$$

and

$${e^{ - 1}} = 1 - {1 \over {1!}} + {1 \over {2!}} - {1 \over {3!}} + .........$$

$$\therefore$$ $$e + {e^{ - 1}} = 2\left[ {1 + {1 \over {2!}} + {1 \over {4!}} + ......} \right]$$

$$\therefore$$ $${1 \over {2!}} + {1 \over {4!}} + {1 \over {6!}} + .......$$

$$ = {{e + {e^{ - 1}}} \over 2} - 1$$

$$ = {{{e^2} + 1 - 2e} \over {2e}}$$

$$ = {{{{\left( {e - 1} \right)}^2}} \over {2e}}$$