### 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)b

^{2}+ c^{2}– 2bc = ac – a^{2}– bc + abb

^{2}+ c^{2}+ a^{2}= ac + bc + ab -(1)Now,

(a + b + c)

^{2}= a^{2}+ b^{2}+ c^{2}+ 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 = ar

^{2}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 = ar^{3}10

^{th}term = ar^{9}16

^{th}term = ar^{15}So, ar

^{9}= = ar^{9}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

b

^{2}= ad -(2)Now,

(a – b)

^{2}= a^{2}+ b^{2}– 2ab= a

^{2}+ ad – a(a + c)From eq(1) and (2), we get

= a

^{2}+ ad – a^{2}– 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 GPR =

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 + b

^{2}+ bc = 2b^{2}+ bc + bab

^{2}+ ac = 2b^{2}b

^{2}= acHence, 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.

c

^{2}= 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.

c

^{2}= aeNow,

c

^{2}(c + e) = ace + c^{2}ec

^{3 }+ c^{2}e = ace + c^{2}ec

^{3 }= acec

^{2 }= aeHence, 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 x^{2}, b^{2}, y^{2} 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.

y

^{2}= bc -(3)Now

2b

^{2}= x^{2}+ y^{2}= (ab) + (bc) -(By using eq(2) and (3))

2b

^{2}= b(a + c)2b

^{2}= b(2b) -(By using eq(1))2b

^{2}= 2b^{2}L.H.S = R.H.S

2b

^{2}= x^{2}+ y^{2}Hence, x

^{2}, b^{2}, y^{2}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.

b

^{2}= ad -(2)Now

(a – b)

^{2}= a(d – c) -(By using eq(2))a

^{2}– 2ab = -aca

^{2}– 2ab = ab – aca(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 = ar

^{2}a + b + c = xb

a + ar + ar

^{2}= x(ar)a(1 + r + r

^{2}) = x(ar)r

^{2}+ (1 – x)r + 1 = 0Here, r is real, so

D ≥ 0

(1 – x)

^{2}– 4(1)(1) ≥ 01 + x

^{2}-2x – 4 ≥ 0x

^{2}– 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, xR

^{2},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}= x

^{0}.R^{0}= 1.1

= 1