Arithmatic Progression
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3
Let the terms be a-d, a, a +d
According to first condition,
a - d + a + a + d = 18
=> 3a = 18
=> a = 6
Now,
According to second condition,
(a-d) ^2 + a^2 + (a+d)^2 = 140
=> ( 6 - d) ^2 + 6^2 + (6+d)^2 = 140
=> 36 + d^2 - 12d + 36 + 36 + d^2 + 12d = 140
=> 108 + 2d^2 = 140
=> 2 d^2 = 140 - 108
=> 2 d^2 = 32
=> d^2 = 16
=> d = 4
Terms are 2, 6, 10
According to first condition,
a - d + a + a + d = 18
=> 3a = 18
=> a = 6
Now,
According to second condition,
(a-d) ^2 + a^2 + (a+d)^2 = 140
=> ( 6 - d) ^2 + 6^2 + (6+d)^2 = 140
=> 36 + d^2 - 12d + 36 + 36 + d^2 + 12d = 140
=> 108 + 2d^2 = 140
=> 2 d^2 = 140 - 108
=> 2 d^2 = 32
=> d^2 = 16
=> d = 4
Terms are 2, 6, 10
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Answered by
21
Let the terms be a-d, a, a +d
According to first condition,
a - d + a + a + d = 18
=> 3a = 18
=> a = 6
Now,
According to second condition,
(a-d) ^2 + a^2 + (a+d)^2 = 140
=> ( 6 - d) ^2 + 6^2 + (6+d)^2 = 140
=> 36 + d^2 - 12d + 36 + 36 + d^2 + 12d = 140
=> 108 + 2d^2 = 140
=> 2 d^2 = 140 - 108
=> 2 d^2 = 32
=> d^2 = 16
=> d = 4
Terms are 2, 6, 10
hope it helped ❤️
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