In the previous section, we completed a discussion on sequences and series. In this section, we will see some miscellaneous examples.
Solved example 9.21
If pth, qth, rth and sth terms of an A.P are in G.P, then show that (p-q), (q-r), (r-s) are also in G.P.
Solution:
1. We want to prove that (p-q), (q-r), (r-s) are in G.P.
• That means, we have to prove that $\frac{q-r}{p-q}~=~\frac{r-s}{q-r}$
2. Let 'a' be the first term and 'd' the common difference of the A.P.
• Then the given terms can be written as:
♦ pth term = ap = a + (p-1)d
♦ qth term = aq = a + (q-1)d
♦ rth term = ar = a + (r-1)d
♦ sth term = as = a + (s-1)d
3. Given that, the above four terms are in G.P. So we can write:
$\begin{array}{ll}
{}&{\frac{a_q}{a_p}}
&{~=~\frac{a_r}{a_q}}& {~=~\frac{a_s}{a_r}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\frac{a+(q-1)d}{a+(p-1)d}}
&{~=~\frac{a+(r-1)d}{a+(q-1)d}}& {~=~\frac{a+(s-1)d}{a+(r-1)d}}
&{\color {green} {\text{- - - - (a)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\frac{a+(q-1)d~-~[a+(p-1)d]}{a+(p-1)d}}
&{~=~\frac{a+(r-1)d~-~[a+(q-1)d]}{a+(q-1)d}}& {~=~\frac{a+(s-1)d~-~[a+(r-1)d]}{a+(r-1)d}}
&{\color {green} {\text{- - - - (b)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\frac{a+qd-d-a-pd+d}{a+(p-1)d}}
&{~=~\frac{a+rd-d-a-qd+d}{a+(q-1)d}}& {~=~\frac{a+sd-d-a-rd+d}{a+(r-1)d}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\frac{(q-p)d}{a+(p-1)d}}
&{~=~\frac{(r-q)d}{a+(q-1)d}}& {~=~\frac{(s-r)d}{a+(r-1)d}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
\end{array}$
◼ Remarks:
(i) Line marked as (a):
This line is obtained by substituting the results from step (2)
(ii) Line marked as (b):
Here we apply componendo and dividendo rule:
$\begin{array}{ll}
{\text{If}}&{\frac{a}{b}~=~\frac{c}{d}}
&{\text{Then}}& {\frac{a-b}{b}~=~\frac{c-d}{d}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\text{Example:}}&{\frac{5}{2}~=~\frac{10}{4}}
&{\text{Then}}& {\frac{5-2}{2}~=~\frac{10-4}{4}~=~\frac{6}{4}~=~\frac{3}{2}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
\end{array}$
4. Consider the above result in (3).
Take out the first two items. We get:
$\begin{array}{ll}
{}&{\frac{(q-p)d}{a+(p-1)d}}
&{~=~\frac{(r-q)d}{a+(q-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\frac{p-q}{a+(p-1)d}}
&{~=~\frac{q-r}{a+(q-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\frac{q-r}{p-q}}
&{~=~\frac{a+(q-1)d}{a+(p-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
\end{array}$
5. Consider again the result in (3).
Take out the last two items. We get:
$\begin{array}{ll}
{}&{\frac{(r-q)d}{a+(q-1)d}}
&{~=~\frac{(s-r)d}{a+(r-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\frac{q-r}{a+(q-1)d}}
&{~=~\frac{r-s}{a+(r-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\frac{r-s}{q-r}}
&{~=~\frac{a+(r-1)d}{a+(q-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
\end{array}$
• From the line marked as (a) in step (3), we get:
$\frac{a+(r-1)d}{a+(q-1)d}~=~\frac{a+(q-1)d}{a+(p-1)d}$
• So we can write: $\frac{r-s}{q-r}~=~\frac{a+(q-1)d}{a+(p-1)d}$
6. Comparing the results in (4) and (5), we get:
$\frac{q-r}{p-q}~=~\frac{a+(q-1)d}{a+(p-1)d}~=~\frac{r-s}{q-r}$
• Hence the statement in (1) is proved.
Solved example 9.22
If a, b, c are in G.P and $a^{\frac{1}{x}}~=~b^{\frac{1}{y}}~=~c^{\frac{1}{z}}$, prove that x, y, z are in A.P.
Solution:
1. We want to prove that x, y, z are in A.P.
• That means, we have to prove that y - x = z - y
2. Given that $a^{\frac{1}{x}}~=~b^{\frac{1}{y}}~=~c^{\frac{1}{z}}$
Let us assume that, the three quantities are equal to k. So we can write:
$a^{\frac{1}{x}}~=~b^{\frac{1}{y}}~=~c^{\frac{1}{z}}~=~k$
3. Consider the above result in (2). Taking the first and last items, we get:
$\begin{array}{ll}
{}&{a^{\frac{1}{x}}}
&{~=~k}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\left(a^{\frac{1}{x}} \right)^x}
&{~=~k^x}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\left(a^{\frac{x}{x}} \right)}
&{~=~k^x}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\left(a^{1} \right)}
&{~=~k^x}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{a}
&{~=~k^x}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
\end{array}$
◼ In a similar way, we will get:
b = ky and c = kz
4. Now we get:
$\frac{b}{a}~=~\frac{k^y}{k^x}~=~k^{y-x}$
5. Also we get:
$\frac{c}{b}~=~\frac{k^z}{k^y}~=~k^{z-y}$
6. Given that a, b and c are in G.P. So we can write: $\frac{b}{a}~=~\frac{c}{b}$
7. So we can equate the results in (4) and (5). We get:
$\frac{b}{a}~=~\frac{c}{b}~=~k^{y-x}~=~k^{z-y}$
• Equating the powers of k, we get: y-x = z-y
• Hence the statement in (1) is proved.
Solved example 9.23
If a, b, c, d and p are different real numbers such that
$\left(a^2 + b^2 + c^2 \right)p^2~-~2(ab+bc+cd)p~+~\left(b^2 + c^2 + d^2 \right) ~\le~0$,
then show that a, b, c and d are in G.P.
Solution:
1. We want to prove that a, b, c and d are in G.P.
• That means, we have to prove that $\frac{b}{a}~=~\frac{c}{b}~=~\frac{d}{c}$.
2. The given inequality can be rearranged as follows:
$\begin{array}{ll}
{}&{\left(a^2 + b^2 + c^2 \right)p^2~-~2(ab+bc+cd)p~+~\left(b^2 + c^2 + d^2 \right)}
&{~\le~0}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\left(a^2 p^2 - 2abp + b^2 \right)~+~\left(b^2 p^2 - 2bcp + c^2 \right)~+~\left(c^2 p^2 - 2cdp + d^2 \right)}
&{~\le~0}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{(ap-b)^2~+~(bp-c)^2~+~(cp-d)^2}
&{~\ge~0}& {\color {green} {\text{- - - - (a)}}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
\end{array}$
◼ Remarks:
Line marked as (a):
• This line is a sum of three terms. Each of the three terms is a square.
♦ Given that: a, b, c, d and p are real numbers.
♦ So (ap-b), (bp-c) and (cp-d) will also be real numbers.
♦ Square of real numbers will be zero or +ve.
♦ So the sum of the three terms will be zero or +ve.
♦ Thus the sign in this line is changed to ≥.
3. The given inequality is only rearranged in the above step (2).
• The rearranged result in (2) is same as the given inequality.
• But the given inequality is ≤. In step (2), it changed to ≥.
• If both ≤ and ≥ are to be satisfied, the only possibility is that, both the expressions are equal to zero.
4. So we can write: $(ap-b)^2~+~(bp-c)^2~+~(cp-d)^2~=~0$
• If the sum of certain squares is zero, each term in that sum must be zero.
• So we get:
(i) (ap-b)2 = 0
(ii) (bp-c)2 = 0
(iii) (cp-d)2 = 0
5. If square of a real number is zero, then that real number must be zero.
So we get:
(i) ap-b = 0
(ii) bp-c = 0
(iii) cp-d = 0
• From 5(i), we get: $p=\frac{b}{a}$
• From 5(ii), we get: $p=\frac{c}{b}$
• From 5(iiii), we get: $p=\frac{d}{c}$
6. Based on the above step (5), we can write:
$\frac{b}{a}~=~\frac{c}{b}~=~\frac{d}{c}~=~p$
• Hence the statement in (1) is proved.
Solved example 9.24
If p, q, r are in G.P and the equations px2 + 2qx + r = 0 and dx2 + 2ex + f = 0 have a common root, then show that $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ are in A.P.
Solution:
1. We want to prove that $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ are in A.P.
• That means, we have to prove that:
$\begin{array}{ll}
{}&{\frac{e}{q} - \frac{d}{p}}
&{~=~\frac{f}{r} - \frac{e}{q}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\frac{e}{q} + \frac{e}{q}}
&{~=~\frac{f}{r} + \frac{d}{p}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\frac{2e}{q}}
&{~=~\frac{f}{r} + \frac{d}{p}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
\end{array}$
2. Given that, p, q, r are in G.P. So we can write: q2 = pr
3. Also given that, px2 + 2qx + r = 0 and dx2 + 2ex + f = 0 have a common root.
• Let us write the roots of the first equation. We get:
$\begin{array}{ll}
{}&{\frac{-2q \pm \sqrt{(2q)^2 ~-~4pr}}{2p}}
&{~=~\frac{-2q \pm \sqrt{4q^2 ~-~4pr}}{2p}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{~=~}&{\frac{-2q \pm \sqrt{4pr ~-~4pr}}{2p}}
&{~=~\frac{-2q \pm \sqrt{0}}{2p}}& {\color {green} {\text{- - - - (a)}}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{~=~}&{\frac{-2q}{2p}}
&{~=~\frac{-q}{p}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
\end{array}$
◼ Remarks:
Line marked as (a): From (2), we have: q2 = pr
4. So the first equation has only one root, which is: $\frac{-q}{p}$
• Given that, the two equations have a common root. So $\frac{-q}{p}$must be a root of the second equation also.
• Since $\frac{-q}{p}$ is a root of the second equation, we can write:
$\begin{array}{ll}
{}&{d \left(\frac{-q}{p} \right)^2~+~2e\left(\frac{-q}{p} \right)~+~f}
&{~=~0}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\frac{d q^2}{p^2}~-~\frac{2eq}{p}~+~f}
&{~=~0}& {}
&{\color {green} {\text{- - - - (a)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\frac{d q^2}{p^2 q^2}~-~\frac{2eq}{p q^2}~+~\frac{f}{q^2}}
&{~=~\frac{d}{p^2}~-~\frac{2e}{p q}~+~\frac{f}{q^2}}& {~=~0}
&{\color {green} {\text{- - - - (b)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\frac{p d}{p^2}~-~\frac{2ep}{p q}~+~\frac{f p}{q^2}}
&{~=~\frac{d}{p}~-~\frac{2e}{q}~+~\frac{f p}{q^2}}& {~=~0}
&{\color {green} {\text{- - - - (c)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\frac{d}{p}~-~\frac{2e}{q}~+~\frac{f p}{pr}}
&{~=~\frac{d}{p}~-~\frac{2e}{q}~+~\frac{f}{r}}& {~=~0}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
{\Rightarrow}&{\frac{2e}{q}}
&{~=~\frac{d}{p}~+~\frac{f}{r}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\
\end{array}$
◼ Remarks:
(i) Line marked as (a):
We want to transform the statement in (a) to the staement in (1). For that, we divide by q2
(ii) Line marked as (b):
We want to further transform the statement in (b) to the statement in (1). For that, we multiply by p.
(iii) Line marked as (c):
From (2), we have: q2 = pr
• Hence the statement in (1) is proved.
The link below gives some more solved examples
Miscellaneous Exercise on chapter 9
In the next chapter we will see straight lines.
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