# Class 11 RD Sharma Solutions – Chapter 19 Arithmetic Progressions- Exercise 19.5

**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 + a

^{2 }– b^{2}– bc / ab = ab + b^{2}− c^{2 }–ac / bcNow, Let us consider the L.H.S. and Multiplying numerator and denominator by ‘c’

we get,

ac

^{2}+ a^{2}c – b^{2}c – bc^{2}/ abcc(b-a)(a+b+c)/abc —(i)

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

a

^{2}b + b^{2}a − c^{2}a – a^{2}c / abca(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 a**^{2 }, b^{2}, c^{2}, are in A.P. Prove that a/(b+c), b/(c+a), c/(a+b) are in A.P.

^{2 }, b

^{2}, c

^{2}, 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 a

^{2}, b^{2}, c^{2 }are in A.P. then, b^{2}– a^{2}= c^{2}– b^{2}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)

b

^{2}+ bc – ac – a^{2}/(a+c)(b+c) = c^{2}+ ac – ab – b^{2}/(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) a ^{2}(b+c), b^{2}(c+a), c^{2}(a+b) are also in A.P.**

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

**(iii) bc-a ^{2}, ca-b^{2}, ab-c^{2} are in A.P.**

**Solution: **

(i) a^{2}(b+c), b^{2}(c+a), c^{2}(a+b) are also in A.P.As we know that, b

^{2}(c+a) – a^{2}(b+c) = c^{2}(a+b) – b^{2}(c+a)b

^{2}c + b^{2}a – a^{2}b – a^{2}c = c^{2}a + c^{2}b – b^{2}c – b^{2}ac(b

^{2}– a^{2}) + ab(b – a) = a(c^{2}– b^{2}) + 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-a^{2}, ca-b^{2}, ab-c^{2}are in A.P.As we know that,

(ca – b

^{2}) – (bc – a^{2}) = (ab – c^{2}) – (ca – b^{2})(ca – b

^{2}– bc + a^{2}) = (ab – c^{2}– ca + b^{2})(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 denominator

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 denominator

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 + a

^{2}– b^{2}– bc/ab = ab + b^{2}– c^{2}– ac/bc(a-b)(a+b+c)/ab = (b-c)(a +b +c)/bc

multiplying with ‘c’ and ‘a’ on numerator and denominator on L.H.S and R.H.S. respectively;

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) a ^{2} + c^{2} + 4ac = 2(ab+bc+ca)**

**(iii) a ^{3} + c^{3} + 6abc = 8b^{3}**

**Solution: **

(i) (a-c)^{2}= 4(a-b)(b-c)Expanding the equation both side;

a

^{2}+ c^{2}– 2ac = 4(ab-ac-b^{2}+bc)a

^{2}+ c^{2}– 2ac = 4ab – 4ac – 4b^{2}+ 4bca

^{2}+ c^{2}+ 4b^{2}+ 2ac + 4bc – 4ab = 0(a + c – 2b)

^{2 }= 0 [Using Identity: (a+b+c)^{2}= a^{2}+b^{2}+c^{2}+2ab+2ac+2bc)](a+c -2b) = 0

a+c -b-b = 0

c-b = b-a

^{3}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) a^{2}+ c^{2}+ 4ac = 2(ab+bc+ca)Expanding the equation both the side;

a

^{2 }+ c^{2}+ 4ac = 2ab+2bc+2caa

^{2}+ c^{2}+ 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, a

^{2}+ c^{2}+ 4ac = 2(ab+bc+ca)

(iii) a^{3}+ c^{3}+ 6abc = 8b^{3}a

^{3}+ c^{3}+ 6abc – (2b)^{3}= 0a

^{3}+ (-2b)^{3}+ c^{3}+ 3 × a × (-2b) × c = 0 [Using identity: x^{3}+y^{3}+z^{3}+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 x**^{2}+xy+y^{2}, z^{2}+zx+x^{2} and y^{2}+yz+z^{2} are in consecutive terms of an A.P. if x,y and z are in A.P.

^{2}+xy+y

^{2}, z

^{2}+zx+x

^{2}and y

^{2}+yz+z

^{2}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 common difference then,

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

Now, (z

^{2}+zx+x^{2})-(x^{2}+xy+y^{2})=(y^{2}+yz+z^{2})-(z^{2}+zx+x^{2})taking L.H.S.

=(z

^{2}+zx+x^{2})-(x^{2}+xy+y^{2})=z

^{2}+zx-xy-y^{2}putting the value of y and z,

=(x+2d)

^{2}+(x+2d)(x)-(x)(x+d)-(x+d)^{2}= x

^{2}+4d^{2}+4xd+x^{2}+2dx-x^{2}-xd-x^{2}-d^{2}-2xd=3xd+3d

^{2}Now, taking R.H.S.

=(y

^{2}+yz+z^{2})-(z^{2}+zx+x^{2})=y

^{2}+yz-zx-x^{2}putting the value of y and z,

=(x+d)

^{2}+(x+d)(x+2d)-(x+2d)(x)-x^{2}=x

^{2}+d^{2}+2xd+x^{2}+2dx+dx+2d^{2}-x^{2}-2dx-x^{2}=3xd+3d

^{2}=L.H.S.

R.H.S=L.H.S

Hence, x

^{2}+xy+y^{2}, z^{2}+zx+x^{2}and y^{2}+yz+z^{2}are in consecutive terms of an A.P.

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