# $\dpi{100} If\; a_{1},a_{2}....,a_{n}\; are\; in\; H.P.,$ then the expression $\dpi{100} a_{1}a_{2}+a_{2}a_{3}+.....+a_{n-1}a_{n}\; is\; equal\; to$ Option 1) $n(a_{1}-a_{n})\;$ Option 2) $\; \; (n-1)(a_{1}-a_{n})\; \;$ Option 3) $\; na_{1}a_{n}\;$ Option 4) $\; (n-1)a_{1}a_{n}$

As we learnt in

Harmonic Progression (HP) -

The sequence $a_{1},a_{2},-------a_{n},$

Where $a_{1}\neq 0$ for each is an HP, if the sequence  $\frac{1}{a_{1}},\frac{1}{a_{2}}-----\frac{1}{a_{n}}$

is an AP

- wherein

eg $\frac{1}{2},\frac{1}{5},\frac{1}{8},\frac{1}{11}----$

If a1, a2,a3,--- are in A.P

Now $\frac{1}{a_{2}}-\frac{1}{a_{1}}=d$

$a_{1}-a_{2}=da_{1}a_{2}$

$\frac{1}{a_{3}}-\frac{1}{a_{2}}=d$

similarly a n-1 -a n=dan-1an

Now a1-a2=da1a2

a2-a3=da2a3

.............................

Add a1 - an = d(a1 a2 + a2 a3+ ..............an - 1 an)                        (i)

Now, $\frac{1}{a_{n}}=\frac{1}{a_{1}}+(n-1)d$

a1-an=(n-1)a1and                                                                            (ii)

So from (i) & (ii) on comparison

Option 1)

$n(a_{1}-a_{n})\;$

this is incorrect option

Option 2)

$\; \; (n-1)(a_{1}-a_{n})\; \;$

this is incorrect option

Option 3)

$\; na_{1}a_{n}\;$

this is incorrect option

Option 4)

$\; (n-1)a_{1}a_{n}$

this is correct option

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