RD Chapter 19 Arithmetic Progressions Ex 19.5 Solutions
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(c + a), c(a + b) are in A.P.
Answer - 1 : -
(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 arein 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 arein AP
1/b – 1/a = 1/c – 1/b
C (b – a) = a (b-c)
Hence, the given termsare 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 weget,
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 inAP
So, 1/a – 1/b = 1/b –1/c
Or c(b-a) = a(c-b)
Hence, given terms arein AP.
Question - 2 : - If a2, b2, c2 are in AP., provethat a/(b+c), b/(c+a), c/(a+b) are in AP.
Answer - 2 : -
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 onboth the sides we get,
Since, b2 –a2 = c2 – b2
Substituting b2 –a2= c2 – b2 in above, we get
LHS = RHS
Hence, given terms arein AP
Question - 3 : - If a, b, c are in A.P., then show that:
(i) a2(b + c), b2(c + a), c2(a + b) are alsoin A.P.
(ii) b + c – a, c + a – b, a + b – c are in A.P.
(iii) bc – a2, ca – b2, ab – c2 are inA.P.
Answer - 3 : -
(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 arein 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 istrue]
Hence, given terms arein 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 LHSand 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 arein AP,
b – a = c – b
Hence, given terms arein 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 LHSand 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 arein AP,
b – c = a – b
Hence, given terms arein AP
Question - 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
Answer - 4 : -
(i) 1/a, 1/b, 1/c are in AP
If 1/a, 1/b, 1/c arein 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 [bymultiplying with ‘c’ on both the numerator and denominator]
Let us consider RHS:
1/c – 1/b = (b-c)/bc
= a(b-c)/bc [bymultiplying 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 inAP then,
ca – bc = ab – ca
c (a-b) = a (b-c)
If 1/a, 1/b, 1/c arein AP then,
1/b – 1/a = 1/c – 1/b
c (a-b) = a (b-c)
Hence, the given termsare in AP
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
Answer - 5 : -
(i) (a – c)2 =4 (a – b) (b – c)
Let us expand theabove 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 inAP
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 theabove 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 inAP
b – a = c – b
b = (a+c)/2
Hence, a2 +c2 + 4ac = 2 (ab + bc + ca)
(iii) a3 + c3 +6abc = 8b3
Let us expand theabove 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 inAP
a – b = c – b
b = (a+c)/2
Hence, a3 +c3 + 6abc = 8b3
Question - 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.
Answer - 6 : -
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 foreach 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 arein AP
Let us multiplynumerator with ‘abc’, we get
abc/bc, abc/ac, abc/abare in AP
∴ a, b, c are in AP.
Hence proved.
Question - 7 : - Show that x2 + xy + y2, z2 +zx + x2 and y2 + yz + z2 are inconsecutive terms of an A.P., if x, y and z are in A.P.
Answer - 7 : -
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 = commondifference,
So, Y = x + d and x =x + 2d
Let us consider theLHS:
(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 considerRHS:
(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.