Question

find the missing term of each arithmetic sequence


1) -7, ___, ___, ___, -43
2) ___, 1/4, ___, ___, 5/4
3) x, ___, ___, ___, ___, ___, 37x​

Answer 1

Solution :-

• a(n) = a + (n - 1)d

1) -7, ___, ___, ___, -43

→ a1 = (-7)

→ a4 = (-43)

→ a4 = a + (4 - 1)d

→ a + 3d = (-43)

→ (-7) + 3d = (-43)

→ 3d = (-43) + 7

→ 3d = (-36)

→ d = (-12)

So,

→ a2 = a + d = (-7) + (-12) = (-19)

→ a3 = a2 + d = (-19) + (-12) = (-31)

2) ___, 1/4, ___, ___, 5/4

→ a2 = a + d = 1/4 --------- (1)

→ a5 = a + 4d = 5/4 -------- (2)

(2) - (1),

→ (a + 4d) - (a + d) = 5/4 - 1/4

→ 3d = 1

→ d = (1/3)

So,

→ a1 = (1/4) - (1/3) = (-1/12)

→ a3 = (-1/12) + (2/3) = (7/12)

→ a4 = (-1/12) + 1 = (11/12)

3) x, ___, ___, ___, ___, ___, 37x

→ a1 = x

→ a7 = a1 + 6d

→ a1 + 6d = 37x

→ x + 6d = 37x

→ 6d = 36x

→ d = 6x .

So,

→ a2 = a + d = x + 6x = 7x

→ a3 = a + 2d = x + 12x = 13x

→ a4 = a + 3d = x + 18x = 19x

→ a5 = a + 4d = x + 24x = 25x

→ a6 = a + 5d = x + 30x = 31x

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