If x=\sum_{n=0}^{\infty }a^{n},y=\sum_{n=0}^{\infty }b^{n},z=\sum_{n=0}^{\infty }c^{n}  where a,b,c  are in A.P. and

\left | a \right |< 1,\left | b \right |< 1,\left | c \right |< 1\; then\; x,y,z\; are\; in

  • Option 1)

    HP\;

  • Option 2)

    Arithmetic­Geometric progression

  • Option 3)

    \; AP\;

  • Option 4)

    \; GP

 

Answers (1)

As we learnt in 

Sum of infinite terms of a GP -

a+ar+ar^{2}+- - - - -= \frac{a}{1-r}\\here \left | r \right |<1

- wherein

a\rightarrow first term

r\rightarrow common ratio

 

Properties of an A.P. -

If a fixed number is added to (Subtracted from) each term of a given A.P, then the resulting sequence is also an AP.

- wherein

If a_{1},a_{2},a_{3},--------- is an AP

then  a_{1}\pm K,a_{2}\pm K,a_{3}\pm K,---------

is also an AP

 

 x=\sum_{n=0}^{\infty}a^{n},\:y=\sum_{n=0}^{\infty}b^{n},\: z=\sum_{n=0}^{\infty}c^{n} 

and a, b, c re in A.P

\therefore x = 1+a+a+...\infty

\Rightarrow     x=\frac{1}{1-a}=1-a=\frac{1}{x}

Y= 1+b+b+...\infty

\Rightarrow    y=\frac{1}{1-b}=1-b=\frac{1}{y}

z=1+c+c+...\infty 

\Rightarrow    z=\frac{1}{1-c}=1-c=\frac{1}{z}

Now, since a, b, c are in A.P

so -a, -b, -c are in A.P

so 1-a, 1-b, 1-c, are in A.P

so \frac{1}{x},\:\frac{1}{y}, \frac{1}{z} in A.P

so x, y, z are in H.P


Option 1)

HP\;

this option is correct

Option 2)

Arithmetic­Geometric progression

this option is incorrect

Option 3)

\; AP\;

this option is incorrect

Option 4)

\; GP

this option is incorrect

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