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Class 11 RD Sharma Solutions – Chapter 19 Arithmetic Progressions- Exercise 19.5
  • Last Updated : 13 Jan, 2021

Question 1. If 1/a, 1/b, 1/c are in A.P. Prove that:

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

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

Solution:  

(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.

Question 2. If a2 , b2, c2, are in A.P. Prove that a/(b+c), b/(c+a), c/(a+b) are in A.P.

Solution: 

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.

Question 3. If a, b, c are in A.P. then show that:

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

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

(iii) bc-a2, ca-b2, ab-c2 are in A.P.

Solution: 

(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.

Question 4. If (b+c)/a, (c+a)/b, (a+b)/c are in A.P. Prove that:

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

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

Solution:

(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.

Question 5. If a, b, c are in A.P. Prove that:

(i) (a-c)2 = 4(a-b)(b-c)

(ii) a2 + c2 + 4ac = 2(ab+bc+ca)

(iii) a3 + c3 + 6abc = 8b3

Solution:  

(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) = 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-b2 = 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.

Question 6. If a(1/b+1/c), b(1/c+1/a), c(1/a+1/b) are in A.P. Prove that a,b,c are in A.P.

Solution: 

(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.

Question 7. Show that x2+xy+y2, z2+zx+x2 and y2+yz+z2 are in consecutive terms of an A.P. if x,y and z are in A.P.

Solution: 

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.

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