In the following questions, two statements (i) and (ii) are given. Choose the valid statement. i. In an A.P., T12 = 37, d = 3 then a = 4. ii. If 1 + 6 + 11 + ... + x = 189 then x = 41.

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Both (i) and (ii) Explanation: (i) Given: T12 = 37, d = 3, n = 12 T12 = a + (n − 1)d 37 = a + (12 − 1)3 37 = a + (11)3 37 = a + 33 37 − 33 = a a = 4 (ii) a = 1, d = 5 Sn = 189 S_n = n/2[2a + (n - 1)d] 189 = n/2[2(1) + (n - 1)5] 189 $\times$ 2 = n[2 + 5n − 5] 378 = n[5n − 3] 5n2 − 3n − 378 = 0 Using the discriminant method: n = `(3 $\pm$ sqrt((-3)^2 - 4(5)(-378)))/10 = (3 $\pm$ sqrt(9 + 7560))/10 = (3 $\pm$ 87)/10 = $\frac{90}{10}$` = 9 x = T9 $\Rightarrow$ T9 = a + (9 − 1)d x = 1 + 8(5) x = 1 + 40 x = 41

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