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ap consits of 37 terms.the sum of the three middle most term is 225 and sum of last 3 terms is 429.find ap
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Let the three middle most terms of the AP be: a – d, a, a + d.
So,
(a – d) + a + (a + d) = 225
3a = 225
a = 75
Now, the AP is
a – 18d, …, a – 2d, a – d, a, a + d, a + 2d, …, a + 18d
Sum of last three terms:
(a + 18d) + (a + 17d) + (a + 16d) = 429
3a + 51d = 429
a + 17d = 143
75 + 17d = 143
d = 4
Now, first term = a – 18d = 75 – 18(4) = 3
Therefore, The AP is 3, 7, 11, … , 147
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So,
(a – d) + a + (a + d) = 225
3a = 225
a = 75
Now, the AP is
a – 18d, …, a – 2d, a – d, a, a + d, a + 2d, …, a + 18d
Sum of last three terms:
(a + 18d) + (a + 17d) + (a + 16d) = 429
3a + 51d = 429
a + 17d = 143
75 + 17d = 143
d = 4
Now, first term = a – 18d = 75 – 18(4) = 3
Therefore, The AP is 3, 7, 11, … , 147
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