⇒ −−
−−−
−−
−−−
()()
()()()
, ()( )
()()()
abac ,
abbcca
bcba
abbcca
()()
()()()
cacb
abbcca
−−
−−−^
are in A.P.
⇒ −
−
−
−
−
−
111
bccaab
,, areinA.P.
⇒
−− −
111
bccaab
,, areinA.P.
- 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 - 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
- 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.
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
- 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
- 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