1.Algebra Booster

Sequence and Series 1.59
Now,
S
n
n
r
r = 1
n
3
 r
r = 1
n
3
= Ât
= ( -1)
Â
= r -
1
r= 1 r=
1
2
Ênn
( + 1) ˆ
= Á -n
Ë
˜
2 ¯
n
Â
Hence, the result.
3. Let t r
= (n – r + 1).r
=nr – r 2 + r
= (n + 1)r – r 2
Now,
S
n
n
= Ât
r
r = 1
n
Â
= (( n + 1) r - r )
r = 1
n
Â
= ( n + 1) r - r
n
Â
2
r= 1 r=
1
Ênn ( + 1) ˆ Ênn ( + 1)(2n+
1) ˆ
= ( n + 1) Á -
Ë
˜
2 ¯ Á
Ë
˜
6 ¯
nn ( + 1) Ê (2n+
1) ˆ
= Á( n + 1) -
2 Ë
˜
3 ¯
nn ( + 1)
= [3( n + 1) - (2 n + 1)]
6
nn ( + 1)
= (3 n + 3 – 2 n – 1)
6
nn ( + 1)( n+
2)
=
6
Hence, the result.
4. Let b = ar, c = ar 2 , where r is the common ratio. It is
given that
a + b + c = xb
fi (a + ar + ar 2 ) = ar.x
fi (1 + r + r 2 ) = r.x
fi r 2 + (1 – x)r + 1 = 0
Since r is real, so D ≥ 0
fi (1 – x) 2 – 4 ≥ 0
fi (x – 1) 2 – 4 ≥ 0
fi (x – 1) 2 – 2 2 ≥ 0
fi (x – 1 + 2)(x – 2) ≥ 0
fi (x + 1)(x – 3) ≥ 0
fi x £ –1 or x ≥ 3
Hence, the result.
5. Given a, b, c are in GP.
\ b 2 = ac
Also, ax 2 + 2bx + c = 0
fi
2
ax + 2 ac x + c = 0
2
fi
2
( ax + c) = 0
fi ( ax + c) = 0
c
fi x =-
a
Since both the given equations have a common root, so,
Êcˆ Ê cˆ
dÁ - 2e + f = 0
Ë
˜ Á ˜
a¯ Ë a¯
fi
2
Êcˆ Ê c ˆ
dÁ - 2eÁ
˜ + f = 0
Ë
˜
a¯
Ë ac¯
Ê
fi c ˆ c
dÁ
- 2e + f = 0
Ë
˜
a¯
b
fi d - 2 e + f = 0
a b c
fi d + f = 2
e
a c b
d e f
\ , , are in AP
a b c
Hence, the result.
6. Given a 1
, a 2
, a 3
, a 4
, …, a n
are in HP.
1 1 1 1
\ , , ,…, are in AP.
a1 a2 a3
a n
fi
1 1 1 1 1 1
– = – =º= – = d
a a a a a a
2 1 3 2 n–1
1 1 a1-
a2
Now, – = d fi = d
a a a a
2 1 1 2
a - a
d
1 2
fi aa 1 2=
a3-
a2
Similarly, aa 2 3=
d
a4-
a3
aa 3 4=
d
o
an-
an–1
an–1an=
d
Thus, a 1
a 2
+ a 2
a 3
+ a 3
a 4
+ … + a n – 1
a n
Êa1-
a2
a2-
a a
3
n-1
- anˆ
= Á + + … +
Ë
˜
d d d ¯
1
= ( a1- a2+ a2- a3+ … + an-1-
an
)
d
1
= ( a1
- an
)
d
= (n – 1)a 1
a n
n
(given)