The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th term is 44. Find the first three terms of the AP.
Answers
Explanation:
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Answer:
a1 = - 13
a2 = -13 + 5 = - 8
a3 = -8 + 5 = - 3
Explanation:
a4 + a8 = 24
a6 + a10 = 44
Use formula an = a + (n - 1)d
a4 = a + (4 - 1)d
a4 = a + 3d
a6 = a + (6 - 1)d
a6 = a + 5d
a8 = a + (8 - 1)d
a8 = a + 7d
a10 = a + (10 - 1)d
a10 = a + 9d
a4 + a8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24
Now calculate the value of a and d
2a = 24 - 10d
a = [24 - 10d]/2 - - - - Eq. 1
a6 + a10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44
2a = 44 - 14d
a = [44 - 14d]/2 - - Eq. 2
Now equate equation 3 and 4
[24 - 10d]/2 = [44 - 14d]/2
2(44 - 14d) = 2(24 - 10d)
88 - 28d = 48 - 20d
-28d + 20d = 48 - 88
-8d = - 40
d = 5
Now substitute the value of d in equation 1 or 2
equation 3
a = [24 - 10d]/2
a = [24 - 10 * 5]/2
a = [24 - 50]/2
a = - 26/2
a = - 13
a1 = - 13
a2 = -13 + 5 = - 8
a3 = -8 + 5 = - 3
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