An A.P consist of 50 terms were 3rd terms is 12 and last term is 106 find the 29th term .also find the sum of the 29th term
Answers
Answered by
2
Hi ,
Let a , d are first term and common
difference of an A.P
********************************************
nth term = Last term = a + ( n - 1 )d
an = a + ( n - 1 )d
********************************************
Now ,
It is given that ,
Third term = 12
a + 2d = 12 ------( 1 )
Last term = 106
a + 49d = 106 ---( 2 )
Subtract ( 1 ) from ( 2 ) , we get
47d = 94
d = 2
Substitute d value in equation ( 1 ) ,
We get
a + 2 × 2 = 12
a = 12 - 4
a = 8
29th term = a + 28d
a29 = 8 + 28 × 2
= 8 + 56
= 64
I hope this helps you.
: )
Let a , d are first term and common
difference of an A.P
********************************************
nth term = Last term = a + ( n - 1 )d
an = a + ( n - 1 )d
********************************************
Now ,
It is given that ,
Third term = 12
a + 2d = 12 ------( 1 )
Last term = 106
a + 49d = 106 ---( 2 )
Subtract ( 1 ) from ( 2 ) , we get
47d = 94
d = 2
Substitute d value in equation ( 1 ) ,
We get
a + 2 × 2 = 12
a = 12 - 4
a = 8
29th term = a + 28d
a29 = 8 + 28 × 2
= 8 + 56
= 64
I hope this helps you.
: )
rituthakur52:
itna to aata tha bt i want sum of 29th term
thnks
Answered by
1

Secondary School
Math
Hi ,
Let a , d are first term and common
difference of an A.P
********************************************
nth term = Last term = a + ( n - 1 )d
an = a + ( n - 1 )d
********************************************
Now ,
It is given that ,
Third term = 12
a + 2d = 12 ------( 1 )
Last term = 106
a + 49d = 106 ---( 2 )
Subtract ( 1 ) from ( 2 ) , we get
47d = 94
d = 2
Substitute d value in equation ( 1 ) ,
We get
a + 2 × 2 = 12
a = 12 - 4
a = 8
29th term = a + 28d
a29 = 8 + 28 × 2
= 8 + 56
= 64
I hope this helps you.
: )
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61 votes
THANKS
202
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Report

Panzer786
Genius
Heya !!!
Third term = 12
A + 2D = 12 ---------(1)
It is given that the AP consists 50 term. So the last term of AP should be 50.
50 term = 106
A + 49D = 106 --------(2)
From equation (1) we get,
A + 2D = 12
A = 12-2D ----------(3)
Putting the Value of A in equation (3) we get,
A + 49D = 106
12-2D + 49D = 106
47D = 106-12
47D = 94
D = 94/47
D = 2
Putting the Value of D in equation (3)
A => 12-2D = 12-2 × 2
A = 12-4 = 8
Therefore,
★ 29th term of AP
=> A +28D
=> 8 + 28 × 2
=> 8 + 56
=> 64.
H
OPE IT WILL HELP YOU....... (+_+)
MARK AS BRAINIEST IF FOUND HELPFULL.
Secondary School
Math
Hi ,
Let a , d are first term and common
difference of an A.P
********************************************
nth term = Last term = a + ( n - 1 )d
an = a + ( n - 1 )d
********************************************
Now ,
It is given that ,
Third term = 12
a + 2d = 12 ------( 1 )
Last term = 106
a + 49d = 106 ---( 2 )
Subtract ( 1 ) from ( 2 ) , we get
47d = 94
d = 2
Substitute d value in equation ( 1 ) ,
We get
a + 2 × 2 = 12
a = 12 - 4
a = 8
29th term = a + 28d
a29 = 8 + 28 × 2
= 8 + 56
= 64
I hope this helps you.
: )
4.5
61 votes
THANKS
202
Comments
Report

Panzer786
Genius
Heya !!!
Third term = 12
A + 2D = 12 ---------(1)
It is given that the AP consists 50 term. So the last term of AP should be 50.
50 term = 106
A + 49D = 106 --------(2)
From equation (1) we get,
A + 2D = 12
A = 12-2D ----------(3)
Putting the Value of A in equation (3) we get,
A + 49D = 106
12-2D + 49D = 106
47D = 106-12
47D = 94
D = 94/47
D = 2
Putting the Value of D in equation (3)
A => 12-2D = 12-2 × 2
A = 12-4 = 8
Therefore,
★ 29th term of AP
=> A +28D
=> 8 + 28 × 2
=> 8 + 56
=> 64.
H
OPE IT WILL HELP YOU....... (+_+)
MARK AS BRAINIEST IF FOUND HELPFULL.
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