The sum of four consecutive terms of an A.P. is 2. The sum of 3rd and 4th terms is 11. Find the terms.
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Answer:
a1 = - 7
a2 = -2
a3 = 3
a4 = 8
Step-by-step explanation:
four consecutive term = a + (a+d) + (a+2d) + (a+3d ) =2
= 4a + 6d = 2
= 2a + 3d = 1 --------(1)
The sum of 3rd and 4th terms is 11.
a + 2d + a + 3d = 11
2a + 5d = 11 -------(2)
subtract eq(1) from eq (2)
(2a+5d) - (2a+3d)= 11-1
2d= 10
d = 5
put in eq (1)
2a = -14
a = -7
a1= -7
a2= -2
a3 = -7+ 2×5
a3 = 3
a4 = -7 + 3×5
a4 = 8
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