(i) If a, b, c, d are four distinct positive quantities in A.P., then show that b c>a d
(ii) If a, b, c, d are four distinct positive quantities in G.P., then show that a+d>b+c \sectionSolution
(i) Since a, b, c, d are in A.P., then A.M. > G.M., for the first three terms. Therefore, b>ac Squaring, we get ( Here 2a+c=b) b2>ac
Similarly, for the last three terms
AM>GM c>bd ( Here 2b+d=c) c2>bd Multiplying (1) and (2), we get $\begin{aligned}
& b^{2} c^{2}>(a c)(b d)
\Rightarrow & b c>a d
\text { (ii) } & \text { Since } a, b, c, d \text { are in G.P. }
\end{aligned}$ again A.M. > G.M. for the first three terms
2a+c>b( since ac=b)⇒a+c>2b
Similarly, for the last three terms
2b+d>c⇒b+d>2c
\left(\begin{array}{l} \right.
since bd=c)
\left. \end{array}\right. Adding (3) and (4), we get
(a+c)+(b+d)>2b+2ca+d>b+c
Eample 12 Ifa, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. Then prove that xb−c⋅yc−a⋅za−b=1$$
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Step-by-Step Solution
Step 1: Analyze the properties of A.P. for part (i)
For part (i), we are given that a, b, c, d are four distinct positive quantities in an Arithmetic Progression (A.P.). In an A.P., the middle term is the arithmetic mean of its neighbors. Thus,
b
is the arithmetic mean of
a
and
c
, and
c
is the arithmetic mean of
b
and
d
.
Step 2: Apply AM-GM inequality for part (i)
Since a, b, c are distinct positive quantities, the Arithmetic Mean (AM) is strictly greater than the Geometric Mean (GM). So,
b>ac
, which implies
b2>ac
. Similarly, for b, c, d, we have
c>bd
, which implies
c2>bd
.
Step 3: Multiply the inequalities for part (i)
Multiplying the two inequalities
b2>ac
and
c2>bd
, we get
b2c2>(ac)(bd)
. Since all quantities are positive, we can take the square root of both sides, which simplifies to bc > ad. This proves the first part.
Step 4: Analyze the properties of G.P. for part (ii)
For part (ii), we are given that a, b, c, d are four distinct positive quantities in a Geometric Progression (G.P.). In a G.P., the middle term is the geometric mean of its neighbors. Thus,
b
is the geometric mean of
a
and
c
, and
c
is the geometric mean of
b
and
d
.
Step 5: Apply AM-GM inequality for part (ii)
Since a, b, c are distinct positive quantities, the Arithmetic Mean (AM) is strictly greater than the Geometric Mean (GM). So,
2a+c>ac
. Since
b=ac
, we have
2a+c>b
, which implies a+c > 2b. Similarly, for b, c, d, we have
2b+d>bd
. Since
c=bd
, we get
2b+d>c
, which implies b+d > 2c.
Step 6: Add the inequalities for part (ii)
Adding the two inequalities a+c > 2b and b+d > 2c, we get (a+c) + (b+d) > 2b + 2c. Rearranging the terms, we subtract
b
and
c
from both sides, which simplifies to a+d > b+c. This proves the second part.