1.Algebra Booster

Sequence and Series 1.29
4. Let the first term = a and the common difference = d
1 1
tm
= fi a + ( m- 1) d =
n
n
1 1
tn
= fi a + ( n - 1) d =
m
m
On subtracting, we get
1 1 m-
n
( m- n)
d = - =
n m mn
1
fi d =
mn
1 1
Also, a+ ( m- 1) =
mn n
1
fi a =
mn
Thus, t mn
= a + (mn – 1)d
1 1
= + ( mn -1)
mn mn
1 + ( mn -1)
=
mn
= 1
5. Let the first term = a and the common difference = d
We have t m–1
+ 2t n+1
fi a + md = 2(a + nd)
fi a = (m – 2n)d
Now, t 3m+1
= a + 3md
= (m –2n)d + 3md = 2(2m –n)d
and t m + n + 1
= a + (m + n)d
= (m – 2n)d + (m + n)d
= (2m – n)d
Hence, the result.
6. Let the first term = a and the common difference = d.
We have T p
= q fi a + (p – 1)d = q
and T p + q
= 0 fi a + (p + q – 1)d = 0 (i)
On subtracting, we get
{(p – 1) – (p + q – 1)}d = q
fi d = –1
Put d = –1 in eq.(i), we get
a – (p – 1) = q
fi a = p + q – 1
Now, T q
= a + (q – 1)d
= p + q – 1 – (q – 1)
= p
7. Do yourself.
8. Let the first term = a and the common difference = d
We have
t m – n
+ t m + n
= a + (m – n – 1)d + a + (m + n – 1)d
= 2a + (m – n – 1 + m + n – 1)d
= 2a + (2m – 1)d
= 2(a + (m – 1)d)
= 2t m
9. (i) Let the first term = A and the common difference
= D
It is given that,
t p
= A + (p – 1)D = a
t q
= A + (q – 1)D = b
t r
= A + (r – 1)D = c
Now, a(q – r) + b(r – p) + c(p – q)
= {A + (p – 1)D}(q – r)
+ {A + (q – 1)D}(r – p)
+ {A + (r – 1)D}(p – q)
= A(q – r + r – p + p – q)
– D(q – r + r – p + p – q)
+ {p(q – r) + q(r – p) + r(p – q)}
= 0 + 0 + 0
= 0.
10. We have
a4
2
a =
7 3
a + 3d
2
fi =
a + 6d
3
fi 3a + 9d = 2a + 12d
fi a = 3d
a6
a + 5d
Now, =
a8
a + 7d
3d
+ 5d
=
3d
+ 7d
8d
=
10d
4
=
5
11. Since a, b, c, d and e are in AP, so
a + e = b + d = 2c
We have
a – 4b + 6c – 4d + e = (a + e) – 4(b + d) + 6c
= 2c – 4(2c) + 6c
= 8c – 8c
= 0
12 Do yourself
13. Let ABCD be a quadrilateral in which –A, –B, –C,
–D are in AP.
Let –A = a – 3d, –B = a – d, –C = a + d, –D = a + 3c
where 2d is the common difference.
It is given that 2d = 10, fi d = 5
Clearly, a = 90°
Thus, the angles are
–A = 90° – 15° = 75°
–B = 90° – 5° = 85°
–C = 90° + 5° = 95°
–D = 90° + 15° = 105°
Hence, the angles are
75°, 85°, 95° and 105°
14. Let the roots be a – d, a, a + d
So, a – d + a + a + d = 12
fi 3a = 12
fi a = 4
Also, (a – d) ◊ a ◊ (a + d) = 28