Q1Long answerExercise 5.3
+ 4 + 7 + 10 +...+ x =287 Solution : Here, 1, 4, 7, 10, ..., x form an AP with a = 1, d = 3, an = x We have, an = a + (n – 1)d So, x = 1 + (n – 1) × 3 = 3n – 2 Also, S= (a + l ) So, 287 = (1 + x) = (1 + 3n – 2) or, 574 = n (3n – 1) or, 3n2 – n – 574 = 0
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(A1) → (B 4) (A2) → (B 5) (A3) → (B 1) (A4) → (B 2) 5 3 11 10