1.Algebra Booster

1.66 Algebra Booster
and b + br = 9
b(1 + r) = 9 …(ii)
Also, it is given that,
b = d and a = r
Putting the value of b and r in Eq. (ii), we get
` d + 9d = 9 and 2a + ad = 31
Solving, we get
25 3
a = 2 or and so d = 3 or
2 2
25 79 83
Hence, the AP is 2, 3, 8, 11, … or , , ,…
2 6 6
2 25 625
and the GP is 3, 6, 12, 24, … or , , ,…
3 3 6
35. Let a be the first term and d be the common difference
of the AP
So, a + md, a + nd, a + rd are in GP
fi (a + nd) 2 = (a + md)(a + rd)
fi a 2 + 2a ◊ nd + n 2 d 2 = a 2 + a(m + r)d + mr ◊ d 2
a(2n – m – r) = d(mr – n 2 )
2
a ( mr - n )
=
…(i)
d (2 n-m-r)
Also, m, n, r are in HP. So,
2mr
n =
…(ii)
m+
r
From Eqs (i) and (ii), we get
n 2
( m + r )-
n
a
=
2
d 2 n - ( m + r)
2
nm ( + r) - 2n
=
2[2 n -( m + r)]
n(( m+ r) - 2 n)
=
2[2 n -( m + r)]
n
=-
2
36. Given x = 1 + 3a + 6a 2 + 10a 3 + … …(i)
y = 1 + 4b + 10b 2 + 20b 3 + …
…(ii)
and S = 1 + 3(ab) + 5(ab) 2 + …
…(iii)
Multiplying Eq. (i) by a, we get
ax = a + 3a 2 + 6a 3 + …
…(iv)
Subtracting Eq. (iv) from Eq. (i), we get
x – ax = 1 + 2a + 3a 2 + 4a 3 + …
fi (1 – a)x = 1 + 2a + 3a 2 + 4a 3 + …
fi (1 – a) 2 x = (1 + 2a + 3a 2 + 4a 3 + …)(1 – a)
fi (1 – a) 2 x = (1 + 2a + 3a 2 + 4a 3 + …)
– (a + 2a 2 + 3a 3 + 4a 4 + …)
= 1 + a + a 2 + a 3 + a 4 + …
1
=
1 - a
1
fi x =
3
(1 - a)
fi
Ê
a = Á1
-
Ë x
1
1/3
ˆ
˜
¯
Ê 1 ˆ
Similarly, b =
Á
1 -
1/4
Ë y
˜
¯
Now, multiplying Eq. (iii) by ab, we get
abS = ab + 3(ab) 2 + 5(ab) 3 + …
Subtracting Eq. (vii) from Eq. (iii), we get
(1 – ab)S = 1 + 2ab + 2(ab) 2 + …
2ab
fi (1 - ab) S = 1 + 1– ab
fi
1 + ab
(1 - ab)
S =
1 - ab
1 + ab
fi S =
2
(1 - ab)
From Eqs (v), (vi) and (viii), we get
-1/3 -1/4
1 + (1- x )(1-
y )
fi S =
-1/3 -1/4 2
[1 -(1 - x )(1 - y )]
…(v)
…(vi)
…(vii)
(viii)
37. Let t n
be the first term of the cube n 3 and
S n
be the sum of first terms of each of n 3 .
S n
= 1 + 3 + 7 + 13 + … + t n – 1
+ t n
S n
= 1 + 3 + 7 + 13 + … + t n – 1
+ t n
Subtracting, we get
0 = (1 + 2 + 4 + 6 + … to n terms) – t n
t n
= (1 + 2 + 4 + 6 + … to n terms)
Ên
- 1ˆ
= 1 + 2 Á [2.1 + ( n -1 -1) ◊1]
Ë
˜
2 ¯
= 1 + (n – 1)n
= n 2 – n + 1
Again, let T n
be the last term of n 3 .
Thus, S n
= 1 + 5 + 11 + 19 + … + T n – 1
+ T n
S n
= 1 + 5 + 11 + 19 + … + T n – 1
+ T n
Subtracting, we get
0 = (1 + 4 + 6 + 8 + … to n terms) – T n
T n
= (1 + 4 + 6 + 8 + … to n terms)
Ên
- 1ˆ
= 1 + 2 Á [2.1 + ( n -1 -1) ◊1]
Ë
˜
2 ¯
= 1 + (n – 1)n
= n 2 – n + 1
and rest of the part you try it from mathematical induction.
38. Let r be the common ratio of the given GP.
2ab 2a ◊ ar 2ar
Given 12 = = =
a + b a + ar a + r
fi ar = 6(1 + r) …(i)
2 2
2bc 2ar ◊ ar 2ar
Also, 36 = = =
2
b + c ar + ar 1+
r
fi ar 2 = 18(1 + r) …(ii)