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

  • Last Updated : 21 Feb, 2021

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.

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(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 4th, 10th, and 16th 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 4th term = ar3

10th term = ar9

16th term = ar15

So, ar9\sqrt{(ar^3)(ar^{15})} = ar9

Therefore, 4th, 10th, 16th 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 pth, qth, rth and sth 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: ap, aq, ar, as 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 xa = xb/2 zb/2 = zc, 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 pth, qth and rth terms of a A.P. and G.P. are both a, b and c respectively, show that ab-c bc-a ca-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 = XRp-1, b = xRq-1, c = xRr-1

Hence, ab-c.bc-a.ca-b = (xRp-1)(q-r)D.(xRq-1)(r-p)D.(xRr-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




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