ICOME
    FOLLOW US: facebook twitter instagram youtube

Sequence & Series

Question



Q1. Tp = q , Tq = p , find Tr = ?
(a) 1
(b) p + q
(c) p + q - r
(d) p - q + r

Solution. c
Let a and b be the first term and the common difference respectively, of an A.P.

∴ Tp = a + (p - 1)d = q ----------------(1)

Tq = a + (q - 1) d = p ------------------(2)

Subtracting (2) from (1), we get : (p - q)d = q - p ⇒d = -1.

putting d = -1 in (1) , we get : a = p + q - r

∴ Tr = a + (r - 1)d = (p + q - 1) + (r - 1) - 1 = p + q - r

Ans = p + q - r
Solution

Q2. Divide 20 into four parts which are in A.P . and such that the product of first and fourth is to the product of 2nd and 3rd in the ratio 2:3 . Find 4th term .
(a) 2 or 8
(b) 2
(c) 8
(d) 4 or 9

Solution. a
Let the four parts in A.P. be : a - 3d , a - d, a + d , a + 3d

∴ Their sum = (a - 3d) + (a - d) + (a + d) + (a + 3d) = 20

⇒ 4a = 20 or a = 5

Also , it is given that :
(a - 3d) (a + 3d) /(a - d) (a + d)
=
2/3


a2 - 9d2 /a2 - d2
=
2/3
⇒
25 - 9d2 / 25 - d2
=
2/3


(∵ a = 5)

or 75 - 27d2 = 50 - 2d2 ⇒ 25d2 = 25 ⇒ d = ± 1
Solution

Q3. If
1/b + c
,
1/c + a
,
1/a + b
are in A.P. , then a2, b2 and c2 are in .
(a) H.P
(b) G.P
(c) A.P
(d) None

Solution. c
We are given that
1/b + c
,
1/c + a
,
1/a + b
are in A.P.
Now multipling by (a + b) (b + c) (c + a) to each term , then (a + b) (c + a), (a + b)(b + c) , (b + c) (c + a) will be in A.P.

⇒ a2 + ab + bc + ca + , b2 + ab + bc + ca , c2 + ab + bc + ca are in A.P

Subtracing ab + bc + ca from each term , we notice that a2 , b2 , c2 are in A.P.
Solution

Q4. A man borrows Rs. 8000 at S.I @ 2.76% per annum . It is decided that the principal and interest are to be paid in 10 monthly instalments . If each installment is double of the preceding installment , find the value of the last installment .
(a) 8
(b) 4096
(c) 1909
(d) 2480

Solution. b
S.I on Rs.8000 @ 2.76% p.a for 10 months .

8000 x 2.76 x 10/100 x 2
= 184 .

Total amount = 8000 + 184 = Rs . 8184

Let the first instalment be x , then

∴ x + 2x + 4x + 8x + 16x + 32x + 64x + 128x + 512x + 1024x = 8184 .

or x (20 + 21 + 22 + ......+ 29) =

⇒ x [
29 - 1/2 - 1
] = 8184 ⇒ 1023x = 8184 ⇒ x =
8184/1023
= 8

Thus the first instalment is : x = Rs.8

The last instalment = 29x = 512 x 8 = Rs 4096
Solution

Q5. If the sum of infinite terms in a G.P is 2 and the sum of their squares is 4/3 , find T3 .
(a) 2/3
(b) 1/3
(c) 4
(d) 1/4

Solution. d
S∞ =
a/1 + r
= 2 -----------------(1)

S∞ =
a2/ 1 + r2
=
4/3
-----------------(2)

On solving equation (1) and (2)

we get a = 1 and r = r2

So T3 = ar2 =
1/4
Solution