Ex 8.2, 4

Transcript

Ex 8.2, 4 An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term. (Hint If '𝑎' is the first term and '𝑑' the common difference, then we arrive at the equations 𝑎+2𝑑=12 and 𝑎+49𝑑=106. Solve this pair of linear equations for '𝑎' and '𝑑'.) For an AP, we know that an = a + (n – 1) d Given that 3rd term is 12 a3 = 12 a + (3 – 1)d = 12 a + 2d = 12 Also, Given that last term is 106 Since the AP consists of 50 terms ∴ Last term = 50th term So, we can write a50 = 106 a + (50 – 1) d = 106 a + 49d = 106 Now, our equations are a + 2d = 12 …(1) a + 49d = 106 …(2) Doing (2) – (1) (a + 49d) – (a + 2d) = 106 – 12 a + 49d – a – 2d = 94 47d = 94 d = 94/47 d = 2 Putting d = 2 in (1) a + 2d = 12 a + 2 × 2 = 12 a + 4 = 12 a = 12 – 4 a = 8 Now, we need to find 29th term Finding 29th term Now, 29th term = a29 = a + (n – 1)d Putting values = 8 + (29 – 1) × 2 = 8 + 28 × 2 = 8 + 56 = 64 Hence, 29th term = a29 = 64