find four consecutive terms in AP whose sum is 12 and the sum of 3rd and 4th term is 14
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According to question
Sum of terms = 12
a+a+d+a+2d+a+3d = 12
=> 4a + 6d = 12
=> 2a + 3d = 6 --------(1)
T3 +T4 = 14
=> a+2d +a +3d = 14
=> 2a + 5d = 14 ---------(2)
Subtracting equation 1 from 2,we get
2d = 8
d = 4
a = - 3
Required AP is - 3, 1, 5, 9
Sum of terms = 12
a+a+d+a+2d+a+3d = 12
=> 4a + 6d = 12
=> 2a + 3d = 6 --------(1)
T3 +T4 = 14
=> a+2d +a +3d = 14
=> 2a + 5d = 14 ---------(2)
Subtracting equation 1 from 2,we get
2d = 8
d = 4
a = - 3
Required AP is - 3, 1, 5, 9
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