If a- 1b-c and 1p-q-1r form two A-P-s- then a-q-c are in A-P- if

As a, 1b,c are in A.P. ∴2b=a+c ...(1) Let a,q,c are in A.P. then 2q=a+c ...(2) Given 1p,q,1r are also in A.P., then 1p +1r =2q Using equation (1), we get ...

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Standard XII

Mathematics

Sum of n Terms

If a, 1b,c an...

Question

If $a, \frac{1}{b}, c$ and $\frac{1}{p}, q, \frac{1}{r}$ form two A.P.'s, then $a, q, c$ are in $A . P .$ if

p, b, r

are in

A . P .

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\frac{1}{p}, \frac{1}{b}, \frac{1}{r}

are in

A . P .

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Both of

p, b, r

and

\frac{1}{p}, \frac{1}{b}, \frac{1}{r}

are in

A . P .

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None of above

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Solution

The correct option is B $\frac{1}{p}, \frac{1}{b}, \frac{1}{r}$ are in $A . P .$
As $a, \frac{1}{b}, c$ are in A.P.
$∴ \frac{2}{b} = a + c . . . (1)$

Let $a, q, c$ are in $A . P .$ then
$2 q = a + c . . . (2)$

Given $\frac{1}{p}, q, \frac{1}{r}$ are also in $A . P .$ , then
$\frac{1}{p} + \frac{1}{r} = 2 q$
Using equation $(1)$ , we get
$\Rightarrow \frac{1}{p} + \frac{1}{r} = a + c$
Using equation $(2)$ , we get
$\Rightarrow \frac{1}{p} + \frac{1}{r} = \frac{2}{b}$

Hence $\frac{1}{p}, \frac{1}{b}, \frac{1}{r}$ are in $A . P .$

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If a,1b,c and 1p,q,1r form two A.P.'s, then a,q,c are in A.P. if

The correct option is B 1p,1b,1r are in A.P.As a,1b,c are in A.P. ∴2b=a+c ...(1) Let a,q,c are in A.P. then 2q=a+c ...(2) Given 1p,q,1r are also in A.P., then 1p+1r=2q Using equation (1), we get ⇒1p+1r=a+c Using equation (2), we get ⇒1p+1r=2b Hence 1p,1b,1r are in A.P.

If $a, \frac{1}{b}, c$ and $\frac{1}{p}, q, \frac{1}{r}$ form two A.P.'s, then $a, q, c$ are in $A . P .$ if