AP consist of 50 terms of which 3rd term is 12 and last term is 106 find 29 th term
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n = 50
T3 = 12
=> a + 2d = 12---------(1)
L = 106
=> a +(n-1)d = 106
=> a + (50-1)d = 106
=> a + 49d = 106 -------(2)
On subtracting equation 1 from 2, we get
47d = 94
d = 2
Now,
on putting the value of d in equation 1, we get
a + 4 = 12
=> a = 8
T29 = a +28d
= 8 + 28*2
= 8 + 56
= 64
T3 = 12
=> a + 2d = 12---------(1)
L = 106
=> a +(n-1)d = 106
=> a + (50-1)d = 106
=> a + 49d = 106 -------(2)
On subtracting equation 1 from 2, we get
47d = 94
d = 2
Now,
on putting the value of d in equation 1, we get
a + 4 = 12
=> a = 8
T29 = a +28d
= 8 + 28*2
= 8 + 56
= 64
jaisuryaclasstopper:
thank you
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