4th term of an ap is 3times the 1st and the 7th term exceeds twice the 3term by 1 find the first term and the common difference
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There is an AP, The 4th term is three times the first. So the relation is
3a = a + 3d, or
2a = 3d, or
a = (3/2)d …(1)
The 7th term exceeds twice the third term by 1. The relation is
a+6d = 2(a+2d)+1, or
a + 6d = 2a+4d+1, or
2a-a = 6d-4d-1, or
a = 2d-1 …(2). Equate (1) and (2)
a = (3/2)d = 2d-1, or
3d = 4d-2, or
4d-3d=2, or d = 2 and a = 3.
The first term is 3 and the common difference is 3.
Check: The series of the AP is 3,5,7,9,11,13,15.
The 4th term is 9, which is three times the first term. Correct.
The 7th term is 15, which is twice the third term plus 1. Correct.
3a = a + 3d, or
2a = 3d, or
a = (3/2)d …(1)
The 7th term exceeds twice the third term by 1. The relation is
a+6d = 2(a+2d)+1, or
a + 6d = 2a+4d+1, or
2a-a = 6d-4d-1, or
a = 2d-1 …(2). Equate (1) and (2)
a = (3/2)d = 2d-1, or
3d = 4d-2, or
4d-3d=2, or d = 2 and a = 3.
The first term is 3 and the common difference is 3.
Check: The series of the AP is 3,5,7,9,11,13,15.
The 4th term is 9, which is three times the first term. Correct.
The 7th term is 15, which is twice the third term plus 1. Correct.
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