| Exercise
8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
Exercise 8.3
Question 1
Three numbers are in A.P. Their sum is 27 and the sum of their squares is 275. Find the
numbers.
Sol :Let the three numbers are in AP = a, a + d, a + 2d
According to the question,
The sum of three terms = 27
⇒ a + (a + d) + (a + 2d) = 27
⇒ 3a + 3d = 27
⇒ a + d = 9
⇒ a = 9 – d …(i)
and the sum of their squares = 275
⇒ a2 + (a + d)2 + (a + 2d)2 = 275
⇒ (9 – d)2 + (9)2 + ( 9 – d + 2d)2 = 275 [from(i)]
⇒ 81 + d2 – 18d + 81 + 81 + d2 + 18d = 275
⇒ 243 + 2d2 = 275
⇒ 2d2 = 275 – 243
⇒ 2d2 = 32
⇒ d2 = 16
⇒ d = √16
⇒ d = ±4
Now, if d = 4, then a = 9 – 4 = 5
and if d = – 4, then a = 9 – ( – 4) = 9 + 4 = 13
So, the numbers are →
if a = 5 and d = 4
5, 9, 13
and if a = 13 and d = – 4
13, 9, 5
Question 2
The sum of three numbers in A.P. is 12 and the sum of their cubes is 408. Find the numbers.
Sol :Let the three numbers are in AP = a, a + d, a + 2d
According to the question,
The sum of three terms = 12
⇒ a + (a + d) + (a + 2d) = 12
⇒ 3a + 3d = 12
⇒ a + d = 4
⇒ a = 4 – d …(i)
and the sum of their cubes = 408
⇒ a3 + (a + d)3 + (a + 2d)3 = 408
⇒ (4 – d)3 + (4)3 + ( 4 – d + 2d)3 = 408 [from(i)]
⇒ (4 – d)3 + (4)3 + ( 4 + d)3 = 408
⇒ 64 – d3 + 12d2 – 48d + 64 + 64 + d3 + 12d2 + 48d = 408
⇒ 192 + 24d2 = 408
⇒ 24d2 = 408 – 192
⇒ 24d2 = 216
⇒ d2 = 9
⇒ d = √9
⇒ d = ±3
Now, if d = 3, then a = 4 – 3 = 1
and if d = – 3, then a = 4 – ( – 3) = 4 + 3 = 7
So, the numbers are →
if a = 1 and d = 3
1, 4, 7
and if a = 7 and d = – 3
7, 4, 1
Question 3 A
Divide 15 into three parts which are in A.P. and the sum of their squares is 83.
Sol :Let the middle term = a and the common difference = d
The first term = a – d and the succeeding term = a + d
So, the three parts are a – d, a, a + d
According to the question,
Sum of these three parts = 15
⇒ a – d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5
and the sum of their squares = 83
⇒ (a – d)2 + a2 + (a + d)2 = 83
⇒ (5 – d)2 + (5)2 + ( 5 + d)2 = 83 [from(i)]
⇒ 25 + d2 – 10d + 25 + 25 + d2 + 10d = 83
⇒ 75 + 2d2 = 83
⇒ 2d2 = 83 – 75
⇒ 2d2 = 8
⇒ d2 = 4
⇒ d = √4
⇒ d = ±2
Case: (i) If d = 2, then
a – d = 5 – 2 = 3
a = 5
a + d = 5 + 2 = 7
Hence, the three parts are
3, 5, 7
Case: (ii) If d = – 2, then
a – d = 5 – ( – 2) = 7
a = 5
a + d = 5 + ( – 2) = 3
Hence, the three parts are
7, 5, 3
Question 3 B
Divide 20 into four parts which are in A.P. such that the ratio of the product of the first and fourth
is to the product of the second and third is 2 : 3.
Sol :Let the four parts which are in AP are
(a – 3d), (a – d), (a + d), (a + 3d)
According to question,
The sum of these four parts = 20
⇒(a – 3d) + (a – d) + (a + d) + (a + 3d) = 20
⇒ 4a = 20
⇒ a = 5 …(i)
Now, it is also given that
product of the first and fourth : product of the second and third = 2 : 3
i.e. (a – 3d) × (a + 3d) : (a – d) × (a + d) = 2
: 3
$\Rightarrow \frac{(a-3 d)(a+3 d)}{(a-d)(a+d)}=\frac{2}{3}$
$\Rightarrow \frac{a^{2}-9 d^{2}}{a^{2}-d^{2}}=\frac{2}{3}$ [∵(a – b)(a + b) = a2 – b2 ]
⇒ 3(a2 – 9d2) = 2(a2 – d2)
⇒ 3a2 – 27d2 = 2a2 – 2d2
⇒ 3a2 – 2a2 = – 2d2 + 27d2
⇒ (5)2 = – 2d2 + 27d2 [from (i)]
⇒ 25 = 25d2
⇒ 1 = d2
⇒ d = ±1
Case I: if d = 1 and a = 5
a – 3d = 5 – 3(1) = 5 – 3 = 2
a – d = 5 – 1 = 4
a + d = 5 + 1 = 6
a + 3d = 5 + 3(1) = 5 + 3 = 8
Hence, the four parts are
2, 4, 6, 8
Case II: if d = – 1 and a = 5
a – 3d = 5 – 3( – 1) = 5 + 3 = 8
a – d = 5 – ( – 1) = 5 + 1 = 6
a + d = 5 + ( – 1) = 5 – 1 = 4
a + 3d = 5 + 3( – 1) = 5 – 3 = 2
Hence, the four parts are
8, 4, 6, 2
Let the three numbers are (a – d), a and (a + d)
According to question,
Sum of these three numbers = 21
⇒ a – d + a + a + d = 21
⇒ 3a = 21
⇒ a = 7 …(i)
and it is also given that
Product of these numbers = 231
⇒(a – d) × a × (a + d) = 231
⇒(7 – d) × 7 × (7 + d) = 231
⇒ 7 × (72 – d2) = 231 [∵ (a – b)(a + b) = a2 – b2]
⇒ 7 × (49 – d2) = 231
⇒ 343 – 7d2 = 231
⇒ – 7d2 = 231 – 343
⇒ – 7d2 = – 112
⇒ d2 = 16
⇒ d = √16
⇒ d = ±4
Case I: If d = 4 and a = 7
a – d = 7 – 4 = 3
a = 7
a + d = 7 + 4 = 11
So, the numbers are
3, 7, 11
Case II: If d = – 4 and a = 7
a – d = 7 – ( – 4) = 7 + 4 = 11
a = 7
a + d = 7 + ( – 4) = 7 – 4 = 3
So, the numbers are
11, 7, 3
Let the three numbers are (a – d), a and (a + d)
According to question,
Sum of these three numbers = 3
⇒ a – d + a + a + d = 3
⇒ 3a = 3
⇒ a = 1 …(i)
and it is also given that
Product of these numbers = – 35
⇒(a – d) × a × (a + d) = – 35
⇒(1 – d) × 1 × (1 + d) = – 35
⇒ 1 × (12 – d2) = – 35 [∵ (a – b)(a + b) = a2 – b2]
⇒ 1 × (1 – d2) = – 35
⇒ 1 – d2 = – 35
⇒ – d2 = – 35 – 1
⇒ – d2 = – 36
⇒ d2 = 36
⇒ d = √36
⇒ d = ±6
Case I: If d = 6 and a = 1
a – d = 1 – 6 = – 5
a = 1
a + d = 1 + 6 = 7
So, the numbers are
– 5, 1, 7
Case II: If d = – 6 and a = 1
a – d = 1 – ( – 6) = 1 + 6 = 7
a = 1
a + d = 1 + ( – 6) = 1 – 6 = – 5
So, the numbers are
7, 1, – 5
Given: a + b + c ≠ 0
and $\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ are in AP
$\Rightarrow \frac{(a-3 d)(a+3 d)}{(a-d)(a+d)}=\frac{2}{3}$
$\Rightarrow \frac{a^{2}-9 d^{2}}{a^{2}-d^{2}}=\frac{2}{3}$ [∵(a – b)(a + b) = a2 – b2 ]
⇒ 3(a2 – 9d2) = 2(a2 – d2)
⇒ 3a2 – 27d2 = 2a2 – 2d2
⇒ 3a2 – 2a2 = – 2d2 + 27d2
⇒ (5)2 = – 2d2 + 27d2 [from (i)]
⇒ 25 = 25d2
⇒ 1 = d2
⇒ d = ±1
Case I: if d = 1 and a = 5
a – 3d = 5 – 3(1) = 5 – 3 = 2
a – d = 5 – 1 = 4
a + d = 5 + 1 = 6
a + 3d = 5 + 3(1) = 5 + 3 = 8
Hence, the four parts are
2, 4, 6, 8
Case II: if d = – 1 and a = 5
a – 3d = 5 – 3( – 1) = 5 + 3 = 8
a – d = 5 – ( – 1) = 5 + 1 = 6
a + d = 5 + ( – 1) = 5 – 1 = 4
a + 3d = 5 + 3( – 1) = 5 – 3 = 2
Hence, the four parts are
8, 4, 6, 2
Question 4 A
Sum of three numbers in A.P. is 21 and their product is 231. Find the numbers.
Sol :Let the three numbers are (a – d), a and (a + d)
According to question,
Sum of these three numbers = 21
⇒ a – d + a + a + d = 21
⇒ 3a = 21
⇒ a = 7 …(i)
and it is also given that
Product of these numbers = 231
⇒(a – d) × a × (a + d) = 231
⇒(7 – d) × 7 × (7 + d) = 231
⇒ 7 × (72 – d2) = 231 [∵ (a – b)(a + b) = a2 – b2]
⇒ 7 × (49 – d2) = 231
⇒ 343 – 7d2 = 231
⇒ – 7d2 = 231 – 343
⇒ – 7d2 = – 112
⇒ d2 = 16
⇒ d = √16
⇒ d = ±4
Case I: If d = 4 and a = 7
a – d = 7 – 4 = 3
a = 7
a + d = 7 + 4 = 11
So, the numbers are
3, 7, 11
Case II: If d = – 4 and a = 7
a – d = 7 – ( – 4) = 7 + 4 = 11
a = 7
a + d = 7 + ( – 4) = 7 – 4 = 3
So, the numbers are
11, 7, 3
Question 4 B
Sum of three numbers in A.P. is 3 and their product is — 35. Find the numbers.
Sol :Let the three numbers are (a – d), a and (a + d)
According to question,
Sum of these three numbers = 3
⇒ a – d + a + a + d = 3
⇒ 3a = 3
⇒ a = 1 …(i)
and it is also given that
Product of these numbers = – 35
⇒(a – d) × a × (a + d) = – 35
⇒(1 – d) × 1 × (1 + d) = – 35
⇒ 1 × (12 – d2) = – 35 [∵ (a – b)(a + b) = a2 – b2]
⇒ 1 × (1 – d2) = – 35
⇒ 1 – d2 = – 35
⇒ – d2 = – 35 – 1
⇒ – d2 = – 36
⇒ d2 = 36
⇒ d = √36
⇒ d = ±6
Case I: If d = 6 and a = 1
a – d = 1 – 6 = – 5
a = 1
a + d = 1 + 6 = 7
So, the numbers are
– 5, 1, 7
Case II: If d = – 6 and a = 1
a – d = 1 – ( – 6) = 1 + 6 = 7
a = 1
a + d = 1 + ( – 6) = 1 – 6 = – 5
So, the numbers are
7, 1, – 5
Question 5
If $\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ are
in A.P. and a + b + c ≠ 0, prove that $\frac{1}{b+c},
\frac{1}{c+a}, \frac{1}{a+b}$are in A.P.
Sol :Given: a + b + c ≠ 0
and $\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ are in AP
To Prove: $\frac{1}{b+c},
\frac{1}{c+a}, \frac{1}{a+b}$ are in AP
if $\frac{a+b+c}{b+c}, \frac{a+b+c}{c+a}, \frac{a+b+c}{a+b}$ are in AP
[multiplying each term by a + b + c]
if $\frac{a+b+c}{b+c}, \frac{a+b+c}{c+a}, \frac{a+b+c}{a+b}$ are in AP
[multiplying each term by a + b + c]
i.e.
if $\frac{a}{b+c}+1$,$\frac{b}{c+a}+1$,$\frac{c}{a+b}+1$ are in
AP
which is given to be true
Hence, $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in AP
a2, b2, c2 are in AP
∴ b2 – a2 = c2 – b2
⇒(b – a)(b + a) = (c – b)(c + b)
$\Rightarrow \frac{b-a}{c+b}=\frac{c-b}{b+a}$
$\Rightarrow \frac{b-a}{(c+b)(c+a)}=\frac{c-b}{(b+a)(c+a)}$
$\Rightarrow \frac{b+c-c-a}{(c+b)(c+a)}=\frac{c+a-a-b}{(c+a)(b+a)}$
$\Rightarrow \frac{(b+c)-(c+a)}{(c+b)(c+a)}=\frac{(c+a)-(a+b)}{(c+a)(b+a)}$
$\Rightarrow \frac{(b+c)}{(c+b)(c+a)}-\frac{(c+a)}{(c+b)(c+a)}=\frac{(c+a)}{(c+a)(b+a)}-\frac{(a+b)}{(c+a)(b+a)}$
$\Rightarrow \frac{1}{c+a}-\frac{1}{c+b}=\frac{1}{b+a}-\frac{1}{c+a}$
which is given to be true
Hence, $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in AP
Question 6
If a2, b2,
c2 are in A.P., show that $\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$are in A.P.
Sol :a2, b2, c2 are in AP
∴ b2 – a2 = c2 – b2
⇒(b – a)(b + a) = (c – b)(c + b)
$\Rightarrow \frac{b-a}{c+b}=\frac{c-b}{b+a}$
$\Rightarrow \frac{b-a}{(c+b)(c+a)}=\frac{c-b}{(b+a)(c+a)}$
$\Rightarrow \frac{b+c-c-a}{(c+b)(c+a)}=\frac{c+a-a-b}{(c+a)(b+a)}$
$\Rightarrow \frac{(b+c)-(c+a)}{(c+b)(c+a)}=\frac{(c+a)-(a+b)}{(c+a)(b+a)}$
$\Rightarrow \frac{(b+c)}{(c+b)(c+a)}-\frac{(c+a)}{(c+b)(c+a)}=\frac{(c+a)}{(c+a)(b+a)}-\frac{(a+b)}{(c+a)(b+a)}$
$\Rightarrow \frac{1}{c+a}-\frac{1}{c+b}=\frac{1}{b+a}-\frac{1}{c+a}$
$\Rightarrow \frac{1}{\mathrm{b}+\mathrm{c}},
\frac{1}{\mathrm{c}+\mathrm{a}}, \frac{1}{\mathrm{a}+\mathrm{b}}$are in A.P
$\Rightarrow \frac{a+b+c}{b+c}, \frac{a+b+c}{c+a},
\frac{a+b+c}{a+b}$ are in A.P
$\Rightarrow \frac{a}{b+c}+1, \frac{b}{c+a}+1, \frac{c}{c+a}+1$ are in A.P
$\Rightarrow \frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ are in A.P
Given: a, b, c are in AP
∴ b – a = c – b …(i)
$\Rightarrow \frac{a}{b+c}+1, \frac{b}{c+a}+1, \frac{c}{c+a}+1$ are in A.P
$\Rightarrow \frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ are in A.P
Question 7 A
If a, b, c are in A.P., prove that
$\frac{1}{\mathrm{bc}}, \frac{1}{\mathrm{ca}}, \frac{1}{\mathrm{ab}}$are in A.P.
Sol :$\frac{1}{\mathrm{bc}}, \frac{1}{\mathrm{ca}}, \frac{1}{\mathrm{ab}}$are in A.P.
Given: a, b, c are in AP
∴ b – a = c – b …(i)
To Prove: $\frac{1}{\mathrm{bc}},
\frac{1}{\mathrm{ca}}, \frac{1}{\mathrm{ab}}$ are in AP
$a_{2}-a_{1}=\frac{1}{c a}-\frac{1}{b c}$ $=\frac{b c-c a}{(c a)(b c)}=\frac{c(b-a)}{(c a)(b c)}$
$a_{3}-a_{2}=\frac{1}{a b}-\frac{1}{c a}$ $=\frac{c a-a b}{(c a)(a b)}=\frac{a(c-b)}{(c a)(a b)}$
$\Rightarrow \frac{c(b-a)}{(c a)(b c)}=\frac{a(c-b)}{(c a)(a b)}$
$\Rightarrow \frac{\mathrm{c}(\mathrm{b}-\mathrm{a})}{\mathrm{bc}}=\frac{\mathrm{a}(\mathrm{c}-\mathrm{b})}{\mathrm{ab}}$
$a_{2}-a_{1}=\frac{1}{c a}-\frac{1}{b c}$ $=\frac{b c-c a}{(c a)(b c)}=\frac{c(b-a)}{(c a)(b c)}$
$a_{3}-a_{2}=\frac{1}{a b}-\frac{1}{c a}$ $=\frac{c a-a b}{(c a)(a b)}=\frac{a(c-b)}{(c a)(a b)}$
$\Rightarrow \frac{c(b-a)}{(c a)(b c)}=\frac{a(c-b)}{(c a)(a b)}$
$\Rightarrow \frac{\mathrm{c}(\mathrm{b}-\mathrm{a})}{\mathrm{bc}}=\frac{\mathrm{a}(\mathrm{c}-\mathrm{b})}{\mathrm{ab}}$
⇒ b – a = c – b
∴a, b, c are in AP
$\therefore \frac{1}{\mathrm{bc}}, \frac{1}{\mathrm{ca}}, \frac{1}{\mathrm{ab}}$ are in AP
Given: a, b, c are in AP
Since, a, b, c are in AP, we have a + c = 2b …(i)
Now, (b + c)2 – a2, (c + a)2 – b2, (a + b)2 – c2 will be in A.P
If (b + c – a)(b + c + a), (c + a – b)(c + a + b), (a + b – c)(a + b + c) are in AP
i.e. if b + c – a, c + a – b, a + b – c are in AP
[dividing by (a + b + c)]
if (b + c – a) + (a + b – c) = 2(c + a – b)
if 2b = 2(c + a – b)
if b = c + a – b
if a + c = 2b which is true by (i)
Hence, (b + c)2 – a2, (c + a)2 – b2, (a + b)2 – c2 are in A.P
Given: a, b, c are in AP
Since, a, b, c are in AP, we have a + c = 2b …(i)
To Prove : $\frac{1}{\sqrt{\mathrm{b}}+\sqrt{\mathrm{c}}}, \frac{1}{\sqrt{\mathrm{c}}+\sqrt{\mathrm{a}}}, \frac{1}{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}$ are in AP
⇒$\Rightarrow \frac{1}{\sqrt{\mathrm{c}}+\sqrt{\mathrm{a}}}-\frac{1}{\sqrt{\mathrm{b}}+\sqrt{\mathrm{c}}}=\frac{1}{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}-\frac{1}{\sqrt{\mathrm{c}}+\sqrt{\mathrm{a}}}$
$\Rightarrow \frac{\sqrt{\mathrm{b}}+\sqrt{\mathrm{c}}-\sqrt{\mathrm{c}}-\sqrt{\mathrm{a}}}{(\sqrt{\mathrm{c}}+\sqrt{\mathrm{a}})(\sqrt{\mathrm{b}}+\sqrt{\mathrm{c})}}=\frac{\sqrt{\mathrm{c}}+\sqrt{\mathrm{a}}-\sqrt{\mathrm{a}}-\sqrt{\mathrm{b}}}{(\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}})(\sqrt{\mathrm{c}}+\sqrt{\mathrm{a}})}$
$\Rightarrow \frac{\sqrt{\mathrm{b}}-\sqrt{\mathrm{a}}}{(\sqrt{\mathrm{b}}+\sqrt{\mathrm{c})}}=\frac{\sqrt{\mathrm{c}}-\sqrt{\mathrm{b}}}{(\sqrt{\mathrm{a}}+\sqrt{\mathrm{b})}}$
⇒ (√b – √a)(√b + √a) = (√c – √b)(√c + √b)
⇒ b – a = c – b
⇒ 2b = a + c, which is True ... from (i)
Hence, the result.
Given: $\frac{b+c-a}{a}, \frac{c+a-b}{b}, \frac{a+b-c}{c}$ are in AP
$\therefore \frac{\mathrm{b}+\mathrm{c}-\mathrm{a}}{\mathrm{a}}+\frac{\mathrm{a}+\mathrm{b}-\mathrm{c}}{\mathrm{c}}=2\left(\frac{\mathrm{c}+\mathrm{a}-\mathrm{b}}{\mathrm{b}}\right)$
$\Rightarrow \frac{b}{a}+\frac{c}{a}-1+\frac{a}{c}+\frac{b}{c}-1=\frac{2 c}{b}+\frac{2 a}{b}-2$
$\Rightarrow \frac{b}{a}+\frac{c}{a}+\frac{a}{c}+\frac{b}{c}-\frac{2 c}{b}-\frac{2 a}{b}=0$
∴a, b, c are in AP
$\therefore \frac{1}{\mathrm{bc}}, \frac{1}{\mathrm{ca}}, \frac{1}{\mathrm{ab}}$ are in AP
Question 7 B
If a, b, c are in A.P., prove that
(b + c)2 — a2, (c + a)2 — b2, (a + b)2 — c2 are in A.P.
Sol :(b + c)2 — a2, (c + a)2 — b2, (a + b)2 — c2 are in A.P.
Given: a, b, c are in AP
Since, a, b, c are in AP, we have a + c = 2b …(i)
Now, (b + c)2 – a2, (c + a)2 – b2, (a + b)2 – c2 will be in A.P
If (b + c – a)(b + c + a), (c + a – b)(c + a + b), (a + b – c)(a + b + c) are in AP
i.e. if b + c – a, c + a – b, a + b – c are in AP
[dividing by (a + b + c)]
if (b + c – a) + (a + b – c) = 2(c + a – b)
if 2b = 2(c + a – b)
if b = c + a – b
if a + c = 2b which is true by (i)
Hence, (b + c)2 – a2, (c + a)2 – b2, (a + b)2 – c2 are in A.P
Question 7 C
If a, b, c are in A.P., prove that
$\frac{1}{\sqrt{b}+\sqrt{c}}, \frac{1}{\sqrt{c}+\sqrt{a}}, \frac{1}{\sqrt{a}+\sqrt{b}}$ are in A.P.
Sol :$\frac{1}{\sqrt{b}+\sqrt{c}}, \frac{1}{\sqrt{c}+\sqrt{a}}, \frac{1}{\sqrt{a}+\sqrt{b}}$ are in A.P.
Given: a, b, c are in AP
Since, a, b, c are in AP, we have a + c = 2b …(i)
To Prove : $\frac{1}{\sqrt{\mathrm{b}}+\sqrt{\mathrm{c}}}, \frac{1}{\sqrt{\mathrm{c}}+\sqrt{\mathrm{a}}}, \frac{1}{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}$ are in AP
⇒$\Rightarrow \frac{1}{\sqrt{\mathrm{c}}+\sqrt{\mathrm{a}}}-\frac{1}{\sqrt{\mathrm{b}}+\sqrt{\mathrm{c}}}=\frac{1}{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}-\frac{1}{\sqrt{\mathrm{c}}+\sqrt{\mathrm{a}}}$
$\Rightarrow \frac{\sqrt{\mathrm{b}}+\sqrt{\mathrm{c}}-\sqrt{\mathrm{c}}-\sqrt{\mathrm{a}}}{(\sqrt{\mathrm{c}}+\sqrt{\mathrm{a}})(\sqrt{\mathrm{b}}+\sqrt{\mathrm{c})}}=\frac{\sqrt{\mathrm{c}}+\sqrt{\mathrm{a}}-\sqrt{\mathrm{a}}-\sqrt{\mathrm{b}}}{(\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}})(\sqrt{\mathrm{c}}+\sqrt{\mathrm{a}})}$
$\Rightarrow \frac{\sqrt{\mathrm{b}}-\sqrt{\mathrm{a}}}{(\sqrt{\mathrm{b}}+\sqrt{\mathrm{c})}}=\frac{\sqrt{\mathrm{c}}-\sqrt{\mathrm{b}}}{(\sqrt{\mathrm{a}}+\sqrt{\mathrm{b})}}$
⇒ (√b – √a)(√b + √a) = (√c – √b)(√c + √b)
⇒ b – a = c – b
⇒ 2b = a + c, which is True ... from (i)
Hence, the result.
Question 8
$\frac{b+c-a}{a}, \frac{c+a-b}{b},
\frac{a+b-c}{c}$ are in A.P., show
that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$are in A.P.
provided a + b +
c 
0
Sol :0Given: $\frac{b+c-a}{a}, \frac{c+a-b}{b}, \frac{a+b-c}{c}$ are in AP
$\therefore \frac{\mathrm{b}+\mathrm{c}-\mathrm{a}}{\mathrm{a}}+\frac{\mathrm{a}+\mathrm{b}-\mathrm{c}}{\mathrm{c}}=2\left(\frac{\mathrm{c}+\mathrm{a}-\mathrm{b}}{\mathrm{b}}\right)$
$\Rightarrow \frac{b}{a}+\frac{c}{a}-1+\frac{a}{c}+\frac{b}{c}-1=\frac{2 c}{b}+\frac{2 a}{b}-2$
$\Rightarrow \frac{b}{a}+\frac{c}{a}+\frac{a}{c}+\frac{b}{c}-\frac{2 c}{b}-\frac{2 a}{b}=0$
Taking LCM
⇒b2c + c2b + a2b + ab2 – 2ac2 – 2a2c = 0
⇒ b2c + c2b + a2b + ab2 –ac2 – ac2 – a2c – a2c = 0
⇒(b2c – a2c) + (c2b – ac2) + (a2b – a2c) + (ab2 – ac2) = 0
⇒ c (b – a)(b + a) + c2(b – a) + a2 (b – c) + a(b + c)(b – c) = 0
⇒ c(b – a) {(b + a) + c} + a(b – c) {a + (b + c)} = 0
⇒ (a + b + c){cb – ca + ab – ca} = 0
Given a + b + c ≠ 0
⇒cb – ca + ab – ca = 0
⇒cb – 2ca + ab = 0
$\Rightarrow \frac{1}{a}-\frac{2}{b}+\frac{1}{c}=0$
$\Rightarrow \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{c}}=\frac{2}{\mathrm{b}}$
$\Rightarrow \frac{1}{\mathrm{a}}, \frac{1}{\mathrm{b}}, \frac{1}{\mathrm{c}}$are in AP
Hence Proved
Given: (b – c)2, (c – a)2, (a – b)2 are in A.P
∴ 2(c – a)2 = (b – c)2 + (a – b)2 …(i)
⇒cb – ca + ab – ca = 0
⇒cb – 2ca + ab = 0
$\Rightarrow \frac{1}{a}-\frac{2}{b}+\frac{1}{c}=0$
$\Rightarrow \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{c}}=\frac{2}{\mathrm{b}}$
$\Rightarrow \frac{1}{\mathrm{a}}, \frac{1}{\mathrm{b}}, \frac{1}{\mathrm{c}}$are in AP
Hence Proved
Question 9
If (b – c)2, (c – a)2, (a – b)2 are in
A.P., then show that: $\frac{1}{b-c}, \frac{1}{c-a}, \frac{1}{a-b}$are in
A.P.
[Hint: Add ab + bc + ca — a2 — b2 — c2 to each term or let ∝= b-c, β = c-a, ४=a-b, then ∝+β+४= 0]
Sol :[Hint: Add ab + bc + ca — a2 — b2 — c2 to each term or let ∝= b-c, β = c-a, ४=a-b, then ∝+β+४= 0]
Given: (b – c)2, (c – a)2, (a – b)2 are in A.P
∴ 2(c – a)2 = (b – c)2 + (a – b)2 …(i)
To Prove: $\frac{1}{b-c},
\frac{1}{c-a}, \frac{1}{a-b}$ are in AP
or $\frac{2}{c-a}=\frac{1}{b-c}+\frac{1}{a-b}$
$\Rightarrow \frac{2}{c-a}=\frac{a-b+b-c}{(b-c)(a-b)}$
$\Rightarrow \frac{2}{c-a}=\frac{a-c}{(b-c)(a-b)}$
⇒2(b – c)(a – b) = (a – c)(c – a)
⇒2[ab – b2 – ca + cb] = ac – a2 – c2 + ac
⇒2ab – 2b2 – 2ac + 2cb = 2ac – a2 – c2
⇒ a2 + c2 – 4ac = 2b2 – 2ab – 2cb
or $\frac{2}{c-a}=\frac{1}{b-c}+\frac{1}{a-b}$
$\Rightarrow \frac{2}{c-a}=\frac{a-b+b-c}{(b-c)(a-b)}$
$\Rightarrow \frac{2}{c-a}=\frac{a-c}{(b-c)(a-b)}$
⇒2(b – c)(a – b) = (a – c)(c – a)
⇒2[ab – b2 – ca + cb] = ac – a2 – c2 + ac
⇒2ab – 2b2 – 2ac + 2cb = 2ac – a2 – c2
⇒ a2 + c2 – 4ac = 2b2 – 2ab – 2cb
Adding both sides, a2 +
c2, we get
⇒2(a2 + c2) – 4ac = a2 + b2 – 2ab + c2 + b2– 2cb
⇒ 2 (a – c)2 = ( b – a)2 + (b – c)2 which is true from (i)
∴(b – c)2, (c – a)2, (a – b)2 are in A.P
$\therefore \frac{1}{\mathrm{b}-\mathrm{c}}, \frac{1}{\mathrm{c}-\mathrm{a}}, \frac{1}{\mathrm{a}-\mathrm{b}}$ are in AP
Hence Proved
Given: a, b, c are in AP
∴ a + c = 2b
$\Rightarrow \mathrm{b}=\frac{\mathrm{a}+\mathrm{c}}{2}$ …(i)
⇒2(a2 + c2) – 4ac = a2 + b2 – 2ab + c2 + b2– 2cb
⇒ 2 (a – c)2 = ( b – a)2 + (b – c)2 which is true from (i)
∴(b – c)2, (c – a)2, (a – b)2 are in A.P
$\therefore \frac{1}{\mathrm{b}-\mathrm{c}}, \frac{1}{\mathrm{c}-\mathrm{a}}, \frac{1}{\mathrm{a}-\mathrm{b}}$ are in AP
Hence Proved
Question 10 A
If a, b, c are in A.P., prove that:
(a-c)2 =4(a-b)(b-c)
Sol :(a-c)2 =4(a-b)(b-c)
Given: a, b, c are in AP
∴ a + c = 2b
$\Rightarrow \mathrm{b}=\frac{\mathrm{a}+\mathrm{c}}{2}$ …(i)
Now taking RHS i.e. 4(a – b)(b –
c)
$\Rightarrow 4\left(\mathrm{a}-\frac{\mathrm{a}+\mathrm{c}}{2}\right)\left(\frac{\mathrm{a}+\mathrm{c}}{2}-\mathrm{c}\right)$ [from(i)]
$\Rightarrow 4\left(\frac{2 a-a-c}{2}\right)\left(\frac{a+c-2 c}{2}\right)$
$\Rightarrow 4\left(\frac{a-c}{2}\right)\left(\frac{a-c}{2}\right)$
⇒(a – c)2
= LHS
Hence Proved
Given: a, b, c are in AP
∴ a + c = 2b …(i)
$\Rightarrow \mathrm{b}=\frac{\mathrm{a}+\mathrm{c}}{2}$ …(ii)
$\Rightarrow 4\left(\mathrm{a}-\frac{\mathrm{a}+\mathrm{c}}{2}\right)\left(\frac{\mathrm{a}+\mathrm{c}}{2}-\mathrm{c}\right)$ [from(i)]
$\Rightarrow 4\left(\frac{2 a-a-c}{2}\right)\left(\frac{a+c-2 c}{2}\right)$
$\Rightarrow 4\left(\frac{a-c}{2}\right)\left(\frac{a-c}{2}\right)$
⇒(a – c)2
= LHS
Hence Proved
Question 10 B
If a, b, c are in A.P., prove that:
a3 + c3 + 6abc = 8b3
Sol :a3 + c3 + 6abc = 8b3
Given: a, b, c are in AP
∴ a + c = 2b …(i)
$\Rightarrow \mathrm{b}=\frac{\mathrm{a}+\mathrm{c}}{2}$ …(ii)
Taking Lhs i.e. a3 +
c3 + 6abc
$\Rightarrow \mathrm{a}^{3}+\mathrm{c}^{3}+6 \mathrm{ac}\left(\frac{\mathrm{a}+\mathrm{c}}{2}\right)$ [from (i)]
⇒ a3 + c3 + 3ac(a + c)
⇒ a3 + c3 3a2c + 3ac2
⇒ (a + c)3
⇒ (2b)3 [from (ii)]
= 8b3 = RHS
Hence Proved
Given: a, b, c are in AP
∴ a + c = 2b …(i)
$\Rightarrow \mathrm{b}=\frac{\mathrm{a}+\mathrm{c}}{2}$ …(ii)
$\Rightarrow \mathrm{a}^{3}+\mathrm{c}^{3}+6 \mathrm{ac}\left(\frac{\mathrm{a}+\mathrm{c}}{2}\right)$ [from (i)]
⇒ a3 + c3 + 3ac(a + c)
⇒ a3 + c3 3a2c + 3ac2
⇒ (a + c)3
⇒ (2b)3 [from (ii)]
= 8b3 = RHS
Hence Proved
Question 10 C
If a, b, c are in A.P., prove that:
(a + 2b — c)(2b + c — a)(c + a — b) = 4abc
[Hint: Put b =$\frac{a+c}{2}$ on L.H.S. and R.H.S.]
Sol :(a + 2b — c)(2b + c — a)(c + a — b) = 4abc
[Hint: Put b =$\frac{a+c}{2}$ on L.H.S. and R.H.S.]
Given: a, b, c are in AP
∴ a + c = 2b …(i)
$\Rightarrow \mathrm{b}=\frac{\mathrm{a}+\mathrm{c}}{2}$ …(ii)
Now, taking LHS i.e. (a + 2b — c)(2b + c — a)(c
+ a — b)
$\Rightarrow\left(a+2 \times \frac{a+c}{2}-c\right)\left(2 \times \frac{a+c}{2}+c-a\right)\left(c+a-\frac{a+c}{2}\right)$
[from (ii)]
$\Rightarrow(a+a+c-c)(a+c+c-a)\left(\frac{2 c+2 a-a-c}{2}\right)$
$\Rightarrow(2 \mathrm{a})(2 \mathrm{c})\left(\frac{\mathrm{a}+\mathrm{c}}{2}\right)$
⇒ 4abc
[from (ii)]
= RHS
Hence Proved
$\Rightarrow\left(a+2 \times \frac{a+c}{2}-c\right)\left(2 \times \frac{a+c}{2}+c-a\right)\left(c+a-\frac{a+c}{2}\right)$
[from (ii)]
$\Rightarrow(a+a+c-c)(a+c+c-a)\left(\frac{2 c+2 a-a-c}{2}\right)$
$\Rightarrow(2 \mathrm{a})(2 \mathrm{c})\left(\frac{\mathrm{a}+\mathrm{c}}{2}\right)$
⇒ 4abc
[from (ii)]
= RHS
Hence Proved
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