The relationship between a, p and y 1s A] 1:2:3 [B]21:3 [C] 3:21 [D) 2:3:1
Answers
Answer:
Page No 257:
Question 1:
Show that each of the progressions given below is an AP. Find the first term, common difference and next term of each.
(i) 9, 15, 21, 27, ...
(ii) 11, 6, 1, −4, ...
(iii) −1,
-5
6
,
-2
3
,
-1
2
, ...
(iv)
√
2
,
√
8
,
√
18
,
√
32
, ...
(v)
√
20
,
√
45
,
√
80
,
√
125
, ...
ANSWER:
(i) The given progression 9, 15, 21, 27, ... .
Clearly, 15 − 9 = 21 − 15 = 27 − 21 = 6 (Constant)
Thus, each term differs from its preceding term by 6. So, the given progression is an AP.
First term = 9
Common difference = 6
Next term of the AP = 27 + 6 = 33
(ii) The given progression 11, 6, 1, −4, ... .
Clearly, 6 − 11 = 1 − 6 = −4 − 1 = −5 (Constant)
Thus, each term differs from its preceding term by −5. So, the given progression is an AP.
First term = 11
Common difference = −5
Next term of the AP = −4 + (−5) = −9
(iii) The given progression −1,
-5
6
,
-2
3
,
-1
2
, ...
Clearly,
-5
6
-(-1)=
-2
3
-(
-5
6
)=
-1
2
-(
-2
3
)=
1
6
(Constant)
Thus, each term differs from its preceding term by
1
6
. So, the given progression is an AP.
First term = −1
Common difference =
1
6
Next term of the AP =
-1
2
+
1
6
=
-2
6
=
-1
3
(iv) The given progression
√
2
,
√
8
,
√
18
,
√
32
, ...
This sequence can be re-written as
√
2
, 2
√
2
, 3
√
2
, 4
√
2
, ...
Clearly, 2
√
2
-
√
2
=3
√
2
-2
√
2
=4
√
2
-3
√
2
=
√
2
(Constant)
Thus, each term differs from its preceding term by
√
2
. So, the given progression is an AP.
First term =
√
2
Common difference =
√
2
Next term of the AP = 4
√
2
+
√
2
=5
√
2
=
√
50
(v) The given progression
√
20
,
√
45
,
√
80
,
√
125
, ...
This sequence can be re-written as 2
√
5
, 3
√
5
, 4
√
5
, 5
√
5
, ...
Clearly, 3
√
5
-2
√
5
=4
√
5
-3
√
5
=5
√
5
-4
√
5
=
√
5
(Constant)
Thus, each term differs from its preceding term by
√
5
. So, the given progression is an AP.
First term = 2
√
5
=
√
20
Common difference =
√
5
Next term of the AP = 5
√
5
+
√
5
=6
√
5
=
√
180