11. In an A.P sum of the first 2 term is 5 and sum of the last 2 term is 129. Then
find the sequence.
Answers
Given :
For an A.P
The sum of first two terms = 5
The sum of last two terms = 129
To Find :
The A.P sequence
Solution :
Let The number in A.P be = a , (a+d) , (a+2d) , (a+3d) , .......
where first term = a
common difference = d
According to question :
The sum of first two terms = 5
i.e a + (a+d) = 5
Or, 2 a + d = 5 .... (i)
And
The sum of last two terms = 129
i.e a + (n-2) d+ a + (n-1)d = 129
Or, 2 a + d (2 n - 3 ) = 129 ... (ii)
Subtracting (i) from (ii), we get
[ 2 a + d (2 n - 3 ) ] - [ 2 a + d ] = 129 - 5
Or, (2 a - 2 a ) + d ( 2 n - 3 - 1 ) = 124
Or,, 0 + d ( 2 n - 4 ) = 124
Or, d ( n - 2 ) =
Or, d ( n - 2 ) = 62
Or, n = + 2
Since n has to be a non-zero positive integer.
Therefore, d can take the values 1, 2, 31 or 62
Basis on this,
For d = 1 ,
2 a + d = 5
Or, 2 a + 1 = 5
Or, 2 a = 5 - 1
Or, 2 a = 4
i.e a = = 2
∴, a = 2 and d = 1
So, Sequence , a , (a+d) , (a+2d) , (a+3d) = 2 , (2 + 1) , (2 + 2 ×1 ) ( 2 + 3 ×1)
= 2 , 3 , 4, 5 ,.....
Hence, The sequence are 2 , 3 , 4 , 5 , ............... Answer
Given : in an ap sum of first two terms is 5 and sum of last two terms is 129
To find : Sequence
Solution:
Let say first two terms
a , a + d
and last two terms
L , L - d
a + a + d = 5
L + L - d = 129
Adding both
2a + 2L = 134
=> a + L = 67
2L - 2a - 2d = 124
=> L - a - d = 62
=> L = 62 + a + d
a + (n-1)d = 62 + a + d
=> (n - 2)d = 62
62 = 1 * 62 or 2* 31
4 possible cases
d = 1 using a + a + d = 5 => a = 2
n = 64
Sequence is 2 , 3 , 4 ........................., 64 , 65
d = 2 using a + a + d = 5 => a = 1.5
1.5 , 3.5 , ................................ 63.5 , 65.5
d = 31 => a = - 13
-13 , 18 , 49 , 80
d = 62 => a = - 28.5
-28.5 , 33.5 , 95.5
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