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Stuck here, help me understand: - Sequence and series - JEE Main

If\; a_{1},a_{2}....,a_{n}\; are\; in\; H.P., then the expression 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}

 
Answers (1)
180 Views
V Vakul

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

so answer is (n-1)a1an


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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