The 11th term of an AP is 80 less than the first term. If the second term is 88, find the third term.
Let the first term is ‘a’ and common difference is‘d’. T2 = a + d = 88----(1) According to question T1 –T11 = 80 ⇒ a -(a + 10d) = 80 ⇒ - 10d = 80 d = - 8 From equation (1) a + (- 8) = 88 ⇒ a = 96 Third term T3 = a + 2d ⇒ 96 + 2 × - 8 = 96 – 16 = 80 ∴ Required answer is 80 .
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