Since t7 – t5 = 12, we have 2d = 12, giving d = 6. With third term a + 2d = 16, substituting d gives a = 4. The required AP is 4, 10, 16, 22, ….
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Since t7 – t5 = 12, we have 2d = 12, giving d = 6. With third term a + 2d = 16, substituting d gives a = 4. The required AP is 4, 10, 16, 22, ….