Open In App

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

Last Updated : 03 Dec, 2021
Improve
Improve
Like Article
Like
Save
Share
Report

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 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 + 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 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) 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 common 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.



Previous Article
Next Article

Similar Reads

Class 11 RD Sharma Solutions - Chapter 19 Arithmetic Progressions- Exercise 19.7 | Set 2
Question 11. A man starts repaying a loan as first installment of Rs. 100. If he increases the installment by Rs. 5 every month, what amount he will pay in the 30th installment? Solution: In the first installment, the man pays Rs. 100, so a1 = 100 and in next installment the man pays Rs. 105, so a2 = 105. So we can conclude, common difference, d =
5 min read
Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.6 | Set 3
Question 49. Find the sum of first n odd natural numbers. Solution: First odd natural numbers are 1, 3, 5, 7, . . .2n – 1. First term(a) = 1, common difference(d) = 3 – 1 = 2 and nth term(an) = 2n – 1. Now by using the formula of sum of n terms of an A.P. Sn = n[a + an] / 2 So, = n[1 + 2n – 1] / 2 = 2n2 / 2 = n2 Hence, the sum of first n odd natura
32 min read
Class 11 RD Sharma Solutions- Chapter 19 Arithmetic Progressions- Exercise 19.7 | Set 1
Question 1. A man saved Rs. 16500 in 10 years. In each year after the first he saved Rs. 100 more than he did in the receding year. How much did he save in the first year?2 Solution: Let the money saved by man in the first year be Rs. x then, as given in the problem, his saving increases by Rs. 100 per year so, x + (x + 100)+ (x + 200)+ (x + 300)+
8 min read
Class 11 RD Sharma Solutions- Chapter 19 Arithmetic Progressions- Exercise 19.2 | Set 2
General Formula Tn = a + (n - 1) * d where, Tn is the nth term a is the first term d = common difference Nth term form end Tn = l - (n - 1) * d where, Tn is the nth term l is the first term d = common difference Question 13. If (m + 1)th term of an A.P. is twice the (n + 1)th term, Prove that (3m + 1)th term is twice the (m+ n + 1)th term. Solution
8 min read
Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.2
Problem 1: Show that the sequence defined by an = 5n – 7 is an A.P., find its common difference. Solution: Given: an = 5n – 7 Now putting n = 1, 2, 3, 4,5 we get, a1 = 5.1 – 7 = 5 – 7 = -2 a2 = 5.2 – 7 = 10 – 7 = 3 a3 = 5.3 – 7 = 15 – 7 = 8 a4 = 5.4 – 7 = 20 – 7 = 13 We can see that, a2 – a1 = 3 – (-2) = 5 a3 – a2 = 8 – (3) = 5 a4 – a3 = 13 – (8) =
7 min read
Class 11 RD Sharma Solutions - Chapter 19 Arithmetic Progressions- Exercise 19.6
Question 1: Find A.M. between:(i) 7 and 13 (ii) 12 and -8 (iii) (x - y) and (x + y) Solution: (i) 7 and 13 Let A be the Arithmetic Mean of 7 and 13. Then,7, A and 13 are in A.P. Now,A - 7 = 13 - A 2A = 13 + 7 A = 20/2 = 10 ∴ A.M. = 10 (ii) 12 and -8 Let A be the Arithmetic Mean of 12 and -8. Then, 12, A and -8 are in A.P. Now, A – 12 = – 8 – A 2A =
7 min read
Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.3
Problem 1: For the following arithmetic progressions write the first term a and the common difference d:(i) -5, -1, 3, 7, ………… Solution: Given sequence is -5, -1, 3, 7, ………… ∴ First term is -5 And, common difference = a2 - a1 = -1 - (-5) = 4 ∴ Common difference is 4 (ii) 1/5, 3/5, 5/5, 7/5, …… Solution: Given sequence is 1/5, 3/5, 5/5, 7/5, …… ∴ Fi
10 min read
Class 11 RD Sharma Solutions - Chapter 19 Arithmetic Progressions- Exercise 19.1
Question 1. If the nth term an of a sequence is given by an = n2 - n +1, write down its first five terms. Solution: We have, an = n2 - n + 1 ---(1)Putting value n = 1 in equation (1), we get a1 = (1)2 - 1 + 1 = 1 Putting value n = 2 in equation (1), we get a2 = (2)2 - 2 + 1 = 3 Putting value n = 3 in equation (1), we get a3 = (3)2 - 3 + 1 = 7 Putti
7 min read
Class 11 RD Sharma Solutions - Chapter 19 Arithmetic Progressions- Exercise 19.3
Question 1. The sum of first three terms of an AP is 21 and the product of first and the third term exceed the second term by 6, find three terms. Solution: We are given that the sum of first three terms are 21. Let's suppose these three terms are a-d, a, a+d. So, a-d + a + a+d = 21 3a = 21 a= 7 And we are given (a-d)*(a+d) - a = 6 a2 - d2 -a = 6 P
4 min read
Class 10 RD Sharma Solutions- Chapter 9 Arithmetic Progressions - Exercise 9.5
Question 1: Find the value of x for which (8x + 4), (6x – 2) and (2x + 7) are in A.P. Solution: Since (8x + 4), (6x – 2) and (2x + 7) are in A.P. Now we know that condition for three number being in A.P.- ⇒2*(Middle Term)=(First Term)+(Last Term) ⇒2(6x-2)=(8x+4)+(2x+7) ⇒12x-4=10x+11 ⇒(12x-10x)=11+4 ⇒2x=15 ⇒x=15/2 Mean value of x=15/2. Question 2: I
5 min read