第11类RD Sharma解决方案–第19章算术级数-练习19.5

问题1.如果AP中有1 / a，1 / b，1 / c，则证明：

(i)(b + c)/ a，(c + a)/ b，(a + b)/ c在AP中

(ii)a(b + c)，b(a + c)，c(a + b)在AP中

解决方案：

(i) (b+c)/a, (c+a)/b, (a+b)/c are in A.P.

As We know that,

if a, b, c are in A.P. then, b-a = c-b

 if 1/a, 1/b, 1/c are in A.P. then, 1/b-1/a = 1/c-1/b

Similarly, (c+a)/b-(b+c)/a = (a+b)/c-(c+a)/b 

taking LCM both side;

a(c+a)-(b+c)b/ab = b(a+b)-(c+a)c/bc

ac + a² – b² – bc / ab = ab + b² − c² –ac / bc

Now, Let us consider the L.H.S. and Multiplying numerator and denominator by ‘c’

we get,

ac² + a²c – b²c – bc² / abc 

c(b-a)(a+b+c)/abc  —(i)

Now, Let us consider the R.H.S. and Multiplying numerator and denominator by ‘a’

a²b + b²a − c²a – a²c / abc  

a(b-c)(a+b+c)/abc  —(ii)

On Comparing (i) and (ii);

we get, L.H.S = R.H.S.

i.e. c(b-a) = a(b-c)

Hence, the given terms are in A.P.

(ii) a(b+c), b(a+c), c(a+b) are in A.P.

As we know that, if a(b+c), b(a+c), c(a+b) are in A.P.

then, b(a+c) – a(b+c) = c(a+b) – b(a+c)

now, let us consider L.H.S.

b(a+c) – a(b+c)

after simplification equation becomes; 

ba + ac – ab – ac = c(b-a) —(i)

Similarly, R.H.S. becomes;

ca + bc – ab – bc = a(c-b) —(ii)

On Comparing (i) and (ii)

L.H.S. = R.H.S.

i.e. c(b-a) = a(c-b)

Therefore, the given terms are in A.P.

问题2.如果a ² ，b ² ，c ²处于AP中，请证明a /(b + c)，b /(c + a)，c /(a + b)位于AP中

解决方案：

As we know that, if a², b², c² are in A.P. then, b² – a² = c² – b²

Similarly, b/(c+a) – a/(b+c) = c/(a+b) – b/(c+a)

now, taking L.C.M both side, we get;

b(b+c) – a(c+a)/(b+c)(c+a) = c(c+a) – b(a+b)/(a+b)(c+a)

b² + bc – ac – a²/(a+c)(b+c) = c² + ac – ab – b²/(a+b)(c+a)

(b-a)(a+b+c)/(a+c)(b+c) = (c-a)(a+b+c)/(a+b)(c+a)

(b-a) = (c-a)

Therefore, the given terms are in A.P.

问题3.如果a，b，c位于AP中，则显示：

(i)a ² (b + c)，b ² (c + a)，c ² (a + b)也在AP中

(ii)b + ca，c + ab，a + bc在AP中

(iii)bc-a ² ，ca-b ² ，ab-c ²位于AP中

解决方案：

(i) a²(b+c), b²(c+a), c²(a+b) are also in A.P.

As we know that, b²(c+a) – a²(b+c) = c²(a+b) – b²(c+a)

b²c + b²a – a²b – a²c = c²a + c²b – b²c – b²a

c(b² – a²) + ab(b – a) = a(c² – b²) + bc(c – b)

(b – a)(ab + bc+ ca) = (c – b)(ab + bc+ ca)

On cancelling (ab + bc + ca) from both the sides, we get;

(b – a) = (c – b)

Hence, the given terms are in A.P.

(ii) b+c-a, c+a-b, a+b-c are in A.P.

As we know that, 

(c + a – b) – (b + c – a) = (a + b – c) – (c + a – b)

2a – 2b = 2b – 2c

taking 2 common from both the sides;

b – a = c – a

Since, a, b, c are in A.P.

Hence, the given terms are in A.P.

(iii) bc-a², ca-b², ab-c² are in A.P.

As we know that,

(ca – b²) – (bc – a²) = (ab – c²) – (ca – b²)

(ca – b² – bc + a²) = (ab – c² – ca + b²)

(a – b)(a + b + c) = (b – c)(a + b + c)

On cancelling (a+b+c) from both the sides;

(a – b) = (b – c)

(b – c) = (a – b)

Hence, the given terms are in A.P.

问题4.如果(b + c)/ a，(c + a)/ b，(a + b)/ c在AP中，请证明：

(i)1 / a，1 / b，1 / c在AP中

(ii)BC，CA，AB在AP中

解决方案：

(i) 1/a, 1/b, 1/c are in A.P.

As we know that,

1/b – 1/a = 1/c – 1/b

let us consider L.H.S

1/b – 1/a = a – b/ab

multiplying by ‘c’ on both numerator and denomenator

we get; c(a – b)/abc

now, let us consider R.H.S.

1/c – 1/b = b – c/cb

multiplying by ‘a’ on both numerator and denomenator

we get that, a(b – c)/abc

Since,  (b+c)/a, (c+a)/b, (a+b)/c are in A.P. 

c + a/b – b + c/a = a + b/c – c + a/b

ac + a² – b² – bc/ab = ab + b² – c² – ac/bc

(a-b)(a+b+c)/ab = (b-c)(a +b +c)/bc

multiplying with ‘c’  and ‘a’ on numerator and denomenator on L.H.S and R.H.S. repectively;

we get,

c(a-b) = a(b-c)

L.H.S. = R.H.S.

hence, the given terms are in A.P.

(ii) bc, ca, ab are in A.P.

As we know that if given terms are in A.P. then,

ab – ca = ca – bc

a(b-c) = c(a-b)

Hence, the given terms are in A.P.

问题5.如果a，b，c在AP中，请证明：

(i)(ac) ² = 4(ab)(bc)

(ii)a ² + c ² + 4ac = 2(ab + bc + ca)

(iii)a ³ + c ³ + 6abc = 8b ³

解决方案：

(i) (a-c)² = 4(a-b)(b-c)

Expanding the equation both side;

a² + c² – 2ac = 4(ab-ac-b²+bc)

a² + c² – 2ac = 4ab – 4ac – 4b² + 4bc

a² + c² + 4b² + 2ac + 4bc – 4ab = 0

(a + c – 2b)²  = 0              [Using Identity: (a+b+c)² = a²+b²+c²+2ab+2ac+2bc)]

(a+c -2b) = 0

a+c -b-b = 0

c-b = b-a³

b-a = c-b

Since, a,b,c are in A.P.

after solving the given term we get;

a+c = 2b

Therefore,  (a-c)² = 4(a-b)(b-c).

(ii) a² + c² + 4ac = 2(ab+bc+ca)

Expanding the equation both the side;

a² + c² + 4ac = 2ab+2bc+2ca

a² + c² + 2ac -2ab-2bc = 0

(a+c-b)²-b^{2 =} 0        [Using Indentity: 

a + c -b = b

a+c = 2b

b = a+c/2

Hence,  a² + c² + 4ac = 2(ab+bc+ca)

(iii) a³ + c³ + 6abc = 8b³

a³ + c³ + 6abc – (2b)³ = 0

a³ + (-2b)³ + c³ + 3 × a × (-2b) × c = 0      [Using identity: x³+y³+z³+3xyz = 0, if x+y+z = 0]

(a-2b+c) = 0

a+c = 2b

a-b = b-c

Since, a,b,c are in A.P.

Hence proved.

问题6.如果a(1 / b + 1 / c)，b(1 / c + 1 / a)，c(1 / a + 1 / b)在AP中证明a，b，c在AP中

解决方案：

(1/b+1/c), b(1/c+1/a), c(1/a+1/b) are in A.P.

Adding 1 in the given terms(it does’nt affect the A.P.) outside the bracket;

(1/b+1/c) +1, b(1/c+1/a)+1, c(1/a+1/b)+1 are in A.P.

Now, taking L.C.M;

we get,

(ac+ab+bc)/bc, (ab+bc+ac)/ac, (bc+ac+ab)/ab are in A.P.

1/bc, 1/ac, 1/ab are in A.P.

Multiplying with ‘abc’ on the numerator,

abc/bc, abc/ac, abc/ab are in A.P.

After solving we get,

a, b, c are in A.P.

Hence proved.

问题7：证明如果x，y和z在AP中^{，则x 2} + xy + y ² ，z ² + zx + x ²和y ² + yz + z ^{2在AP的连续项中}

解决方案：

As given that, x , y and z are in A.P.

Let d is the commom difference then,

y = x+d, z = x+2d

Now, (z²+zx+x²)-(x²+xy+y²)=(y²+yz+z²)-(z²+zx+x²)

taking L.H.S.

=(z²+zx+x²)-(x²+xy+y²)

=z²+zx-xy-y²

putting the value of y and z,

=(x+2d)²+(x+2d)(x)-(x)(x+d)-(x+d)²

= x²+4d²+4xd+x²+2dx-x²-xd-x²-d²-2xd

=3xd+3d²

Now, taking R.H.S.

=(y²+yz+z²)-(z²+zx+x²)

=y²+yz-zx-x²

putting the value of y and z,

=(x+d)²+(x+d)(x+2d)-(x+2d)(x)-x²

=x²+d²+2xd+x²+2dx+dx+2d²-x²-2dx-x²

=3xd+3d²

=L.H.S.

R.H.S=L.H.S

Hence, x²+xy+y², z²+zx+x² and y²+yz+z² are in consecutive terms of an A.P.