Either (a – c)^2 = 0 ⇒ a = c

and from (1),

b

=aa+ =aabc⇒ ==

2

Or, (a + c)^2 + 2ac = 0 ⇒ a^2 + c^2 + 4ac = 0

or ac acor ac

b

(^22) ac
22
(^42)
+=−^2 =− 4
[from (2)]
or ac=− 2 b or b =−⋅a c
2
22
Hence −abc
2
,, are in G.P.

18. Let a be the rth mean between x and 2y and b be the
    rth mean between 2x and y.
    Here, n arithmetic means have been inserted in
    both cases.
    ∴ Number of terms of A.P. in each case = n + 2
    Now, a = rth A.M. between x and 2y
    = (r + 1)th term of A.P.
    = x + rd, where d is the common difference of A.P.
    =+ −
    
    
    • 
      
    
    
    
    

⎛

⎝⎜

⎞

⎠⎟

xryx

n

2

1

∵c.d. last term term

no. of terms

st

= −

−

⎡

⎣

⎢

⎢

⎤

⎦

⎥

⎥

1

1

Similarly, bxryx

n

=+ −

+

⎛

⎝⎜

⎞

(^2) ⎠⎟
2
1
According to question, a = b
∴ + −

• ⎛
  ⎝⎜
  ⎞
  ⎠⎟=+
  −


• 
  

  ⎛
  ⎝⎜
  ⎞
  xr ⎠⎟
  yx
  n
  xryx
  n
  2
  1
  2 2
  1
  ⇒ =

  
  


• 
  

  x r −−+
  n
  yxy x
  1
  () 22
  ⇒ =

  
  


• 
  

  x r +
  n
  yx
  1
  ()⇒ (n + 1)x = r(y + x)

  
  

19. Since 1st group contains one number, 2nd group
    contains 2 numbers, 3rd group contains 3 numbers
    and so on, therefore, nth group will contain n
    numbers.
    Here we observe that numbers in each group are in
    A.P. whose c.d. is 1. Therefore, number in nth group
    will be in A.P. having c.d. 1.
    Thus, for nth group, d = 1, n= n
    Sequence of first terms of groups is 2, 4, 7, .....
    First terms of this sequence is the first term of the
    first group.
    Second term is the first term of the second group.
    Third term is the first term of the third group and
    so on.
    nth term of this sequence i.e., tn will be the first term
    of the nth group.

Let Sn = 1 + 2 + 4 + 7 + ..... + tn ...(1)

Sn = 1 + 2 + 4 + ..... + tn–1 + tn ...(2)

Subtracting (2) from (1), we get

0 = 1 + [1 + 2 + 3 + ..... + to (n – 1) terms]–tn

or tn=+ 1 nnnn−^1 = − +

2

2

2

()^2

Thus for the nth group,

a (first term) =nn− + dnn==

(^22)
2
,, 1
∴ Sum of numbers in the nth group

⎛ − +
⎝⎜
⎞
⎠⎟

• −⋅
  ⎧
  ⎨
  ⎩
  ⎫
  ⎬
  ⎭
  n nn n =nn+
  2
  2 2
  2
  11 1
  2
  2 2
  () ()

20. The sequence of the difference between successive
    terms is 2, 6, 18, 54, ....
    This is a G.P. with first term 2 and common ratio 3.
    Let Sn = 5 + 7 + 13 + 31 ..... + tn ...(1)
    Also, Sn = 5 + 7 + 13 + ..... + tn – 1 + tn ...(2)
    Subtracting (2) from (1), we get
    0 = 5 +[2 + 6 + 18 + ..... to (n – 1) terms] –tn
    or tn = 5 + [2 + 6 + 18 + ..... to (n – 1) terms]
    =+ −
    −

=+ − =+

− −−

5231

31

53 143

()^111

()

()

n nn

Now,

Stnrr

n r

r

n

==+

=

−

=

∑∑

1

1

1

() 43 =+

=

−

=

∑∑^43

1

1

r 1

n r

r

n

= 4n + (1 + 3 + 3^2 + ..... + 3n–1)

= + −

−

4 13 1 =+−

31

4 31

2

nn

() ()nn

=+^1 −

2

[] 381 n n

””