In this exercise, students will be solving problems based on the properties of arithmetic progressions. For students who find difficulty in solving problems, the RD Sharma Solutions for Class 11 Maths is the right study material one can opt for. Students can quickly refer to RD Sharma solutions for any doubts pertaining to the problems. Experts at BYJU’S have created the solutions as per the guidelines, in accordance with the latest CBSE patterns, which are easily available in PDF format too. Students can refer to RD Sharma Class 11 Maths Solutions PDF and can download them from the links below.
RD Sharma Solutions for Class 11 Maths Exercise 19.5 Chapter 19 – Arithmetic Progressions
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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(c + a), c(a + b) are in A.P.
Solution:
(i) (b+c)/a, (c+a)/b, (a+b)/c are in A.P.
We know that if a, b, c are in AP, then b – a = c – b
If, 1/a, 1/b, 1/c are in AP
Then, 1/b – 1/a = 1/c – 1/b
If (b+c)/a, (c+a)/b, (a+b)/c are in AP
Then, (c+a)/b – (b+c)/a = (a+b)/c – (c+a)/b
Let us take LCM
Since, 1/a, 1/b, 1/c are in AP
1/b – 1/a = 1/c – 1/b
C (b – a) = a (b-c)
Hence, the given terms are in AP.
(ii) a(b + c), b(c + a), c(a + b) are in A.P.
We know that if, b(c + a) – a(b+c) = c(a+b) – b(c+a)
Consider LHS:
b(c + a) – a(b+c)
Upon simplification, we get,
b(c + a) – a(b+c) = bc + ba – ab – ac
= c (b-a)
Now,
c(a+b) – b(c+a) = ca + cb – bc – ba
= a (c-b)
We know,
1/a, 1/b, 1/c are in AP
So, 1/a – 1/b = 1/b – 1/c
Or c(b-a) = a(c-b)
Hence, the given terms are in AP.
2. If a2, b2, c2 are in AP., prove that a/(b+c), b/(c+a), c/(a+b) are in AP.
Solution:
If a2, b2, c2 are in AP then, b2 – a2 = c2 – b2
If a/(b+c), b/(c+a), c/(a+b) are in AP then,
b/(c+a) – a/(b+c) = c/(a+b) – b/(c+a)
Let us take LCM on both sides we get,
Since, b2 – a2 = c2 – b2
Substituting b2 – a2= c2 – b2 in above, we get
LHS = RHS
Hence, given terms are in AP
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.
If b2(c + a) – a2(b + c) = c2(a + b) – b2(c + a)
b2c + b2a – a2b – a2c = c2a + c2b – b2a – b2c
Given, b – a = c – b
And since a, b, c are in AP,
c(b2 – a2 ) + ab(b – a) = a(c2 – b2 ) + bc(c – b)
(b – a) (ab + bc + ca) = (c – b) (ab + bc + ca)
Upon cancelling, ab + bc + ca from both sides
b – a = c – b
2b = c + a [which is true]
Hence, given terms are in AP
(ii) b + c – a, c + a – b, a + b – c are in A.P.
If (c + a – b) – (b + c – a) = (a + b – c) – (c + a – b)
Then, b + c – a, c + a – b, a + b – c are in A.P.
Let us consider LHS and RHS
(c + a – b) – (b + c – a) = (a + b – c) – (c + a – b)
2a – 2b = 2b – 2c
b – a = c – b
And since a, b, c are in AP,
b – a = c – b
Hence, the given terms are in AP.
(iii) bc – a2, ca – b2, ab – c2 are in A.P.
If (ca – b2) – (bc – a2) = (ab – c2) – (ca – b2)
Then, bc – a2, ca – b2, ab – c2 are in A.P.
Let us consider LHS and RHS
(ca – b2) – (bc – a2) = (ab – c2) – (ca – b2)
(a – b2 – bc + a2) = (ab – c2 – ca + b2)
(a – b) (a + b + c) = (b – c) (a + b + c)
a – b = b – c
And since a, b, c are in AP,
b – c = a – b
Hence, given terms are in AP
4. If (b+c)/a, (c+a)/b, (a+b)/c are in AP., prove that:
(i) 1/a, 1/b, 1/c are in AP
(ii) bc, ca, ab are in AP
Solution:
(i) 1/a, 1/b, 1/c are in AP
If 1/a, 1/b, 1/c are in AP then,
1/b – 1/a = 1/c – 1/b
Let us consider LHS:
1/b – 1/a = (a-b)/ab
= c(a-b)/abc [by multiplying with ‘c’ on both the numerator and denominator]
Let us consider RHS:
1/c – 1/b = (b-c)/bc
= a(b-c)/bc [by multiplying with ‘a’ on both the numerator and denominator]
Since (b+c)/a, (c+a)/b, (a+b)/c are in AP
(ii) bc, ca, ab are in AP
If bc, ca, ab are in AP, then,
ca – bc = ab – ca
c (a-b) = a (b-c)
If 1/a, 1/b, 1/c are in AP then,
1/b – 1/a = 1/c – 1/b
c (a-b) = a (b-c)
Hence, the given terms are in AP
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)
Let us expand the above expression
a2 + c2 – 2ac = 4(ab – ac – b2 + bc)
a2 + 4c2b2 + 2ac – 4ab – 4bc = 0
(a + c – 2b)2 = 0
a + c – 2b = 0
Since a, b, c are in AP
b – a = c – b
a + c – 2b = 0
a + c = 2b
Hence, (a – c)2 = 4 (a – b) (b – c)
(ii) a2 + c2 + 4ac = 2 (ab + bc + ca)
Let us expand the above expression
a2 + c2 + 4ac = 2 (ab + bc + ca)
a2 + c2 + 2ac – 2ab – 2bc = 0
(a + c – b)2 – b2 = 0
a + c – b = b
a + c – 2b = 0
2b = a+c
b = (a+c)/2
Since a, b, c are in AP
b – a = c – b
b = (a+c)/2
Hence, a2 + c2 + 4ac = 2 (ab + bc + ca)
(iii) a3 + c3 + 6abc = 8b3
Let us expand the above expression
a3 + c3 + 6abc = 8b3
a3 + c3 – (2b)3 + 6abc = 0
a3 + (-2b)3 + c3 + 3a(-2b)c = 0
Since, if a + b + c = 0, a3 + b3 + c3 = 3abc
(a – 2b + c)3 = 0
a – 2b + c = 0
a + c = 2b
b = (a+c)/2
Since a, b, c are in AP
a – b = c – b
b = (a+c)/2
Hence, a3 + c3 + 6abc = 8b3
6. If a(1/b + 1/c), b(1/c + 1/a), c(1/a + 1/b) are in AP., prove that a, b, c are in AP.
Solution:
Here, we know a(1/b + 1/c), b(1/c + 1/a), c(1/a + 1/b) are in AP
Also, a(1/b + 1/c) + 1, b(1/c + 1/a) + 1, c(1/a + 1/b) + 1 are in AP
Let us take LCM for each expression then we get,
(ac+ab+bc)/bc , (ab+bc+ac)/ac, (cb+ac+ab)/ab are in AP
1/bc, 1/ac, 1/ab are in AP
Let us multiply the numerator with ‘abc’, and we get
abc/bc, abc/ac, abc/ab are in AP
∴ a, b, c are in AP.
Hence proved.
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:
x, y, z are in AP
Given, x2 + xy + y2, z2 + zx + x2 and y2 + yz + z2 are in AP
(z2 + zx + x2) – (x2 + xy + y2) = (y2 + yz + z2) – (z2 + zx + x2)
Let d = common difference,
So, Y = x + d and x = x + 2d
Let us consider the LHS:
(z2 + zx + x2) – (x2 + xy + y2)
z2 + zx – xy – y2
(x + 2d)2 + (x + 2d)x – x(x + d) – (x + d)2
x2 + 4xd + 4d2 + x2 + 2xd – x2 – xd – x2 – 2xd – d2
3xd + 3d2
Now, let us consider RHS:
(y2 + yz + z2) – (z2 + zx + x2)
y2 + yz – zx – x2
(x + d)2 + (x + d)(x + 2d) – (x + 2d)x – x2
x2 + 2dx + d2 + x2 + 2dx + xd + 2d2 – x2 – 2dx – x2
3xd + 3d2
LHS = RHS
∴ x2 + xy + y2, z2 + zx + x2 and y2 + yz + z2 are in consecutive terms of A.P
Hence proved.
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