in ap the 5th and 9th term is 45 and 81 find 1st term,common difference,sum of 1st 15 terms
Answers
Answer:
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
Step-by-step explanation:
