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:

\frac{a^2+ab+b^2}{bc + ca+ab}=\frac{b+a}{c+b}

Solution: 

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

So, a, b = ar, c = ar²

\frac{a^2+ab+b^2}{bc + ca+ab}=\frac{b+a}{c+b}\\ \frac{a^2+a(ar)+a^2r^2}{(ar)(ar^2)+(ar^2)a+a(ar)}=\frac{ar+a}{ar^2+ar}\\ \frac{a^2(1+r+r^2)}{a^2(r^3+r^2+r)}=\frac{a(1+r)}{a(r^2+r)}\\ \frac{1+r+r^2}{r(1+r+r^2)}=\frac{1+r}{r(1+r)}

1/r = 1/r 

L.H.S = R.H.S

Hence, proved \frac{a^2+ab+b^2}{bc + ca+ab}=\frac{b+a}{c+b}

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 = \sqrt{(ar^3)(ar^{15})} = 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 = \frac{a_q}{a_p}=\frac{a_r}{a_q}\\ =\frac{a_q-a_r}{a_p-a_q}\ \ \ -(Using \ ratio\ property)\\ =\frac{[a+(q-1)d]-[a+(r-1)d]}{[a+(p-1)d]-[a+(q-1)d]}\\ =\frac{(q-r)d}{(p-q)d}\\ R=\frac{q-r}{p-q}\ \ \ \ -(1)

Now,

R=\frac{a_r}{a_q}=\frac{a_s}{a_r}\\ =\frac{a_r-a_s}{a_q-a_r}\ \ \ \ -(Using \ ratio\ property)\\ =\frac{[a+(r-1)d]-[a+(s-1)d]}{[a+(q-1)d]-[a+(r-1)d]}\\ =\frac{(r-s)d}{(q-r)d}\\ R=\frac{r-s}{q-r}\ \ \ \ \  -(2)

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

\frac{q-r}{p-q}=\frac{r-s}{p-r}\\   

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

Question 17. If \frac{1}{a+b},\frac{1}{2b},\frac{1}{b+c}   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.

\frac{2}{2b}=\frac{1}{(a+b)}+\frac{1}{(b+c)}\\ \frac{1}{b}=\frac{b+c+a+b}{(a+b)(b+c)}\\ \frac{1}{b}=\frac{2b+c+a}{ab+ac+b^2+bc}

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:

x^2=x^{\frac{b}{2}}z^{\frac{b}{2}}=z^c=\lambda(say)\\ x=\lambda^{\frac{1}{a}},\ z=\lambda^{\frac{1}{c}}\\ \lambda^{\frac{1}{a}\left(\frac{b}{2}\right)}\times\lambda^{\frac{b}{2}\times\frac{1}{c}}=\lambda\\ \lambda^{\frac{b}{2a}+\frac{b}{2c}}=\lambda^1

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,

c^2=bd=2b\times\frac{d}{2}\\ ⇒ c^2=(a+c)\times\frac{ce}{c+e}\\ ⇒ c^2=\frac{(a+c)ce}{c+e}\ \ \ \ \ \left[\because\frac{2}{d}=\frac{e+c}{ce}\right]

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