Class 11 RD Sharma Solutions – Chapter 20 Geometric Progressions- Exercise 20.5 | Set 2

Question 12. If (a – b), (b – c), (c – a) are in G.P., then prove that (a + b + c)² = 3(ab + bc + ca)

Solution: 

Given: (a – b), (b – c), (c – a) are in G.P.

(b – c)² = (a – b)(c – a)

b² + c² – 2bc = ac – a² – bc + ab

b² + c² + a² = ac + bc + ab         -(1)

Now,

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

= ac + bc + ab + 2ab + 2bc + 2ca

So, using eq(1), we get 

= 3ab + 3bc + 3ca

(a + b + c)² = 3(ab + bc + ca)

LHS = RHS

Hence, proved 

Question 13. If a, b, c are in G.P., then prove that:

Solution: 

Given: a, b, c are in G.P.

So, a, b = ar, c = ar²

1/r = 1/r 

L.H.S = R.H.S

Hence, proved 

Question 14. If the 4^th, 10^th, and 16^th terms of a G.P. are x, y, and z respectively. Prove that x, y, z are in G.P.

Solution: 

Let us considered the 4^th term = ar³

10^th term = ar⁹

16^th term = ar¹⁵

So, ar⁹ =  = ar⁹

Therefore, 4^th, 10^th, 16^th terms are also in G.P.

Hence, proved

Question 15. If a, b, c are in A.P. and a, b, d are in G.P., then prove that a, a – b, d – c are in G.P.

Solution:

Given: a, b, c are in A.P.

2b = a + c          -(1)

also,

a, b, d are in G.P., so

b² = ad         -(2)

Now, 

(a – b)² = a² + b² – 2ab

= a² + ad – a(a + c)

From eq(1) and (2), we get

= a² + ad – a² – ac

= ad – ac

(a – b)² = a(d – c)

(a – b)/a = (d – c)/(a – b)

Hence, proved a, (a – b), (d – c) are in G.P.

Question 16. If p^th, q^th, r^th and s^th terms of an A.P. be in G.P., then prove that p – q, q – r, r – s are in G.P.

Solution:

 Let us considered R be common ratio,

Given: a_p, a_q, a_r, a_s of AP are in GP

R = 

Now,

Using eq(1) and (2), we get

Hence, proved (p – q), (q – r), (r – s) are in G.P.

Question 17. If  are the three consecutive terms of an A.P., prove that a, b, c are the three consecutive terms of a G.P.

Solution:

Given: \frac{1}{a+b},\frac{1}{2b},\frac{1}{b+c}   are in A.P.

ab + ac + b² + bc = 2b² + bc + ba

b² + ac = 2b²

b² = ac

Hence, proved a, b , c are in G.P.

Question 18. If x^a = x^b/2 z^b/2 = z^c, then prove that 1/a, 1/b, 1/c are in A.P.

Solution:

b/2a + b/2c = 1

1/a + 1/c = 2/b 

Hence, 1/a, 1/b, 1/c are in A.P.

Question 19. If a, b, c are in A.P., b, c, d are in G.P. and 1/c, 1/d, 1/e are in A.P., prove that a, c, e are in G.P.

Solution:

Given: a, b, c are in A.P.   

2b = a + c         -(1)

Also, b, c, d are in G.P.   

c² = bd         -(2)

1/c, 1/d, 1/e are in A.P.      

2/d = 1/c + 1/e         -(3)

We need to prove that

a, b, c are in G.P.

c² = ae

Now,

c²(c + e) = ace + c²e

c³ + c²e = ace + c²e

c³ = ace

c² = ae

Hence, proved.

Question 20. If a, b, c are in A.P. and a, x, b and b, y, c are in G.P., show that x², b², y² are in A.P.

Solution:

Given: a, b, c are in A.P.

2b = a + c         -(1)

Also, a, x, b are in G.P.

x = ab         -(2)

and b, y, c are in G.P.

y² = bc         -(3)

Now

2b² = x² + y²

= (ab) + (bc)              -(By using eq(2) and (3))

2b² = b(a + c)

2b² = b(2b)                  -(By using eq(1))

2b² = 2b²

L.H.S = R.H.S

2b² = x² + y²

Hence,  x², b², y² are in A.P.

Question 21. If a, b, c are in A.P. and a, b, d are in G.P., show that a, (a – b), (d – c) are in G.P.

Solution:

Given: a, b, c are in A.P.

2b = a + c         -(1)

Also, a, b, d are in G.P.

b² = ad         -(2)

Now

(a – b)² = a(d – c)        -(By using eq(2))

a² – 2ab = -ac

a² – 2ab = ab – ac

a(a – b) = a(b – c)

a – b = a – c

2b = a + c

a + c = a + c,            -(By using eq(1))

L.H.S = R.H.S

Hence, a, (a – b), (d – c) are in G.P.

Question 22. If a, b, c are three real numbers in G.P. and a + b + c = xb, then prove that either x < -1 or x > 3.

Solution:

Let us considered r be the common ratio of G.P.

So, a, b = ar, c = ar²

a + b + c = xb

a + ar + ar² = x(ar)

a(1 + r + r²) = x(ar)

r² + (1 – x)r + 1 = 0

Here, r is real, so

D ≥ 0

(1 – x)² – 4(1)(1) ≥ 0

1 + x² -2x – 4 ≥ 0

x² – 2x – 3 ≥ 0

(x – 3)(x + 1) ≥ 0

Hence, x < -1 or x > 3

Question 23. If p^th, q^th and r^th terms of a A.P. and G.P. are both a, b and c respectively, show that a^b-c b^c-a c^a-b = 1.

Solution:

Let us considered the A.P. be A, A + D, A + 2D, …. and G.P. be x, xR, xR², 

Then

a = A + (p – 1)D, B = A + (q – 1)D, c = A + (r – 1)D

a – b = (p – q)D, b – c = (q – r)D, c – a = (r – p)D

Also a = XR^p-1, b = xR^q-1, c = xR^r-1

Hence, a^b-c.b^c-a.c^a-b = (xR^p-1)^(q-r)D.(xR^q-1)^(r-p)D.(xR^r-1)^(p-q)D

= x^{(q-r+r-p+p-q)D}.R^{[(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)]D}

= x⁰.R⁰

= 1.1 

= 1