⇒ −−

−−−

−−

−−−

()()

()()()

, ()( )

()()()

abac ,

abbcca

bcba

abbcca

()()

()()()

cacb

abbcca

−−

−−−^

are in A.P.

⇒ −

−

−

−

−

−

111

bccaab

,, areinA.P.

⇒

−− −

111

bccaab

,, areinA.P.

12. Let r be the common ratio of the given G.P.
    Then, b = ar, c = ar^2 and d = ar^3
    Now, L.H.S. = (b – c)^2 + (c – a)^2 + (d – b)^2
    = (ar – ar^2 )^2 + (ar^2 – a)^2 + (ar^3 – ar)^2
    = a^2 r^2 (1 – r)^2 + a^2 (r^2 – 1)^2 + a^2 r^2 (r^2 –1)^2
    = a^2 [r^2 (1 – r)^2 + (r^2 –1)^2 + r^2 (r^2 –1)^2 ]
    = a^2 [r^2 (1 + r^2 – 2r) + (r^4 – 2r^2 + 1) + r^2 (r^4 – 2r^2 + 1)]
    = a^2 [r^2 + r^4 – 2r^3 + r^4 – 2r^2 + 1 + r^6 – 2r^4 + r^2 ]
    = a^2 [r^6 – 2r^3 + 1] = a^2 (r^3 – 1)^2 = [a(r^3 – 1)]^2
    = (ar^3 – a)^2 = (d – a)^2 = (a – d)^2


13. Given, x + y + z = 15 ...(1)
    when a, x, y, z, b are in A.P.
    ∴ Sum of A.M.’s = x + y + z =⎛⎝⎜ab+ ⎞⎠⎟⋅
    2

3

⇒ 15 =⎛⎝⎜ + ⎠⎟⎞⋅ +=

2

ab 310 orab ...(2)

Also,^1115

xyz 3

++= when^11111

ax yzb

,,,,are in A.P.

∴ Sum of A.M.’s =^111

11

2

3

xyz

++=ab

⎛⎝⎜ + ⎞⎠⎟

⋅

or^5

32

3 10

2

=⎛ + 3

⎝⎜

⎞

⎠⎟

⋅ =⎛

⎝⎜

⎞

⎠⎟

ab ⋅

ab ab

[From (2)]

∴ ab = 9

⇒ G.M. of a and bab== 3

14. Given Sn = 5n^2 + 3n
    ∴ Sn–1 = 5(n – 1)^2 + 3(n – 1)
    Now, tn = Sn – Sn – 1, n ≥ 2
    = 5n^2 + 3n – 5(n – 1)^2 – 3(n – 1)
    = 5[n^2 – (n – 1)^2 ] + 3[n – n + 1]
    = 5(2n – 1) + 3 = 10n – 2, n ≥ 2
    ∴ t 1 = S 1 = 5. 12 + 3.1 = 8
    t 2 = 10. 2 – 2 = 18
    t 3 = 10. 3 – 2 = 28 and so on.
    Thus terms of the given series are 8, 18, 28 ..... which
    are in A.P. whose c.d. is 10.


15. 
    

    For general term we will take the rth term.
    Since rth term of the sequence 1, 2, 3, .....
    = 1 + (r – 1). 1 = r
    and rth term of the sequence n, n –1, n – 2, .....
    = n + (r – 1)(–1) = n – r + 1
    Now, rth term (tr) = r(n – r + 1) = nr – r^2 + r
    ∴ ==− +

    
    

St nrrrnr∑∑

r

n

r

n

1

2

1

()

= − +=+ ⎛⎝⎜ − + +⎞⎠⎟

== =

nr r∑∑ ∑r

nn n n

r

n

r

n

r

n

1

2

11

1

2

21

3

() 1

=

+ ⎡ −−+

⎣⎢

⎤

⎦⎥

nn() (^1) nn =nn()( )++n
2
3213
3
12
6

16. Given equations are x^2 – 3x + p = 0 ...(1)
    and x^2 – 12x + q = 0 ...(2)
    Given a, b, c, d are in G.P.
    Let r be its common ratio.
    Then, b = ar, c = ar^2 , d = ar^3
    Now a and b i.e., a and ar are roots of equation (1)
    ∴ a(1+ r) = 3 ...(3)
    and a^2 r = p ...(4)
    Again, c and d i.e., ar^2 and ar^3 are the roots of
    equation (2)
    ∴ ar^2 (1 + r) = 12 ...(5)
    and a^2 r^5 = q ...(6)
    Dividing (5) by (3), we get r^2 = 4 ∴ r = ±2
    Now qp
    qp

ar ar

ar ar

r

r

+

−

= +

−

= +

−

=

25 2

25 2

4

4

1

1

17

15

17. Given a, b, c are in A.P. ∴ b=ac+
    2

...(1)

and^111222

abc

,,

are in A.P.

∴ =+=+ =

+

211 2

222

22

22

2 22

bac^22

ac

ac

b ac

ac

or

...(2)

From (1) and (2), we get ac ac

ac

⎛ +

⎝⎜

⎞

⎠⎟

=

(^2) +
(^2222)
22
⇒ (a + c)^2 (a^2 + c^2 ) = 8a^2 c^2
⇒ (a^2 + c^2 + 2ac)(a^2 + c^2 ) = 8a^2 c^2
⇒ (a^2 + c^2 )^2 + 2ac(a^2 + c^2 ) – 8a^2 c^2 = 0
⇒ {(a^2 + c^2 )^2 – 4a^2 c^2 } + {2ac(a^2 + c^2 ) – 4a^2 c^2 }=0
⇒ (a^2 – c^2 )^2 + 2ac(a^2 + c^2 – 2ac) = 0
⇒ (a^2 – c^2 )^2 + 2ac(a – c)^2 = 0
⇒ (a – c)^2 [(a + c)^2 + 2ac] = 0