Sequence & Series
Question
Q1.
Find an A.P whose 4th term is 9 and 9th term is 19 . Find its 35th term .
Solution.
a + 3d = 9 ------------------(1)
a + 8d = 19 --------------(2)
on solving equation (1) and (2)
d = 3 , a = 3
Tn = a + (n - 1)d
T35 = 3 + 34 x 2 = 71
Solution
Q2.
It 6 times the 6th term of an A.P. is equal to 10 times its 10th term . Find 16th term .
Solution.
6T6 = 10T10
6(a + 5d) = 10(a + 9d)
6a + 30d = 10a + 90d
- 4a = 60 d ⇒ 4a + 60d = 0 ⇒ 4 (a + 15d) = 0
a + 15d = T16
So T16 = 0
Solution
Q3.
If pth term of an A.P is q and the qth term is P . find show the rth term .
Solution.
Tp = q ⇒ a + (p - 1)d = q -----------(1)
Tq = p ⇒ a + (q - 1)d = p -----------(2)
solving equation (1) and (2)
we get d = -1
a = p + q - 1
Tr = a + (r - 1)d = p + q - 1 + (r - 1)(-1) =
p + q - r
Solution
Q4.
Find the sum of the first 10 terms of the series , whose Tn =
n + 10/5
Solution.
Put n = 1 , we get T
1 = a =
11/5
T
2 =
12/5
,
So d =
1/5
S
10 =
10/2
[
22/5
+
9/5
] =
5 [
31/5
] = 31
Solution
Q5.
Find the 6th term of the series whose sum to n terms is n(5x + 2)
Solution.
Tn = Sn - Sn - 1 = 10x - 3
Put n = 6 , T6 = 57
Solution
Q6.
If
1/b + c
,
1/c + a
,
1/a + b
are in A.P then a2
, b2 , c2 are in ____________ .
(a) AP (b) GP (c) HP
(d) None
Solution.
multiply each term by (b + c)(c + a)(a + b) we get (c + a)(a + b),(b + c)(a + b),(b + c)(c + a)
ca + cb + a2 + ab , ba + b2 + ca + cb , bc + ba + c2 + ca
Now substract each term by ab +
bc + ca , we get a2 , b2 , c2 are A.P
Solution
Q7.
If S10 = S15. Find S25
Solution.
10/2
[2a + 9d] =
15/2
[2a + 14d]
2[2a + 9d] = 3[2a + 14d]
4a + 18d = 6a + 42d
0 = 2a + 24d
S
25 =
25/2
[2a + 24d]
So, S
25 = 0
Solution
Q8.
If a , b , c are in A.P and b, c, a are in G.P then
1/c
,
1/a
,
1/b
form an ________ .
(a) H.P (b) G.P (c) A.P (d) None
Solution.
a, b, c are in A.P , so 2b = a + c
b, c, a are in G.P , So c
2 = ab
using both conditions , we get
2/a
=
1/c
+
1/b
which shows A.P
Solution
Q9.
Find the sum of n terms
5 + 55 + 555 + ----------
Solution.
= 5[1 + 11 + 111 + ---------to n terms]
=
5/9
[9 + 99 + 999 + ----- to n terms]
=
5/9
[(10 - 1) + (100 - 1) + (1000 - 1) + -------- to n terms]
=
5/9
[
10(10n - 1)/10 - 1
- n]
=
5/8
[
10
n + 1 - 9x - 10].
Solution
Q10.
In a G.P the sum of infinite terms is 15 , the sum of squares of these infinite terms is 45 . find a and d .
Solution.
S
∞ =
a/1 - r
= 15
---------(1)
S
∞2 =
a2/
1 - r2
= 45---------(2)
squaring equation one
a2/
(1 - r)2
= 225 ---------(3)
Divide equation (3) by equation (2)
1 - r2/(1 - r)2
=
225/45
⇒
r =
2/3
, a = 5
Solution
