18. Find the relation between x and y such that the rth
    mean between x and 2y may be same as the rth mean
    between 2x and y, if n means are inserted in each case.


19. Natural numbers are divided into groups in the
    following way : 1;(2, 3); (4, 5, 6); (7, 8, 9, 10)
    Show that the sum of the numbers in the nth group

is nn()

(^21)
2
+.

20. Find 5 + 7 + 13 + 31 + 85 + ..... to n terms.

SOLUTIONS

1. The terms of given series are in A.P. whose common
   difference d = – 4 and first term a = 99.
   Now, sum of 20 terms of the series,

S^20

20

2

= [299 20 1 4⋅ +( −−)( )]

= 10 (198 – 76) = 1220

2. Let A 1 , A 2 , ....., An be n arithmetic means between
   a and b. Then, a, A 1 , A 2 , ....., An, b is an A.P. with
   common difference d given by d ba
   n

= −

+ 1

.^

Now, AA A

nAA

12 +++= +.... nn 2 ( 1 )

(^) =+nab
2
()=n⎛⎝⎜ab+ ⎟⎞⎠=×nab
2
()A.M. between and

3. (a + b) +(a^2 + 2b) + (a^3 + 3b) + ..... to n terms
   = (a + a^2 + a^3 + ..... to n terms)
   
   
   • b(1 + 2 + 3 + ..... to n terms)
   
   
   
   

= −

−

aa+ ⋅ +

a

b nn

() () 1 n

1

1

2

4. Given,

212

xyyyz++

,, are in A.P.

∴ =

+

+

+

= ++

++

22 2 1 2

yxyyz y

xyz

xyyz

or

()()

or xy + 2y^2 + yz = xy + y^2 + xz + yz

or y^2 = xz ∴ x, y, z in G.P.

5. 
   

   Here a =1, r=^1
   3
   So, S a
   ∞ r

   
   

   −

   
   

=

−

=

1

1

1 1

3

3

2

6. Let 2n be the number of arithmetic means between
   two numbers a and b.

Now sum of the 2n A.M.’s between a and b

=

ab+ ⋅nabn=+

2

2( )

Given, ()abnn+=+ 21 ⇒⋅^13 nn=+

6

21

∴ n = 6. Hence, number of means = 2n = 12

7. Given x, y, z are in A.P.
   ∴ 2 y = x + z ...(1)
   A 1 is the A.M. of x and y ∴ A 1 =xy+
   2

...(2)

A 2 is the A.M. of y and z ∴ A 2 =yz+

2

...(3)

Adding (2) and (3), we get

AA 12 xz y y y^2

2

22

2

+=++ = + = 2y [From (1)]

∴ y=AA^12 +

2

∴ y is the A.M. of A 1 and A 2.

8. Let the numbers be (a – d), a, (a + d). Then,
   (a – d) + a + (a + d) = – 3 ⇒ a = –1
   Also, (a – d)(a)(a + d) = 8 ⇒ a(a^2 – d^2 ) = 8
   ⇒ (–1)(1 – d^2 ) = 8 [... a = –1]
   ⇒ d^2 = 9 ⇒ d = ± 3
   When a = –1 and d = 3, the numbers are –4, –1, 2.
   When a = –1 and d = –3, the numbers are 2, –1, –4.
   So, the numbers are –4, –1, 2 or 2, –1, –4


9. Here, a = 1, r=^1
   2

and n = 10 ∴ (^) Sar
r
10
(^101)
1
= −
−
⎛
⎝⎜
⎞
⎠⎟
⇒ =
⎛
⎝⎜
⎞
⎠⎟ −
⎛
⎝⎜
⎞
⎠⎟−
⎧
⎨
⎪⎪
⎩
⎪
⎪
⎫
⎬
⎪⎪
⎭
⎪
⎪
S 10 = ⎛⎝⎜ − ⎞⎠⎟
10
(^110)
1
2
1
1
2
1
21 1
2
= −
⎛
⎝⎜
⎞
⎠⎟
2 21 = − =
2
1024 1
512
1023
512
10
10
()

10. Since A.M. ≥ G.M.

∴ xx+ ≥⇒x + ≥ 1 ⇒ + ≥

x

xx x

x

1

2

11

2

/. / (^12)

11. a^2 + 2bc, b^2 +2ac, c^2 + 2ab are in A.P.
    ⇒ (a^2 + 2bc) – (ab + bc + ca), (b^2 + 2ac) –(ab + bc + ca),
    (c^2 + 2ab) – (ab + bc + ca) are in A.P.
    ⇒ a^2 + bc – ab – ca, b^2 + ca – ab – bc, c^2 + ab – bc – ca
    are in A.P.
    ⇒ (a – b)(a – c), (b – c)(b – a), (c – a)(c – b) are in A.P.