Saraswati Mathematics Term 2 ILLUSTRATIVE EXAMPLES Example 1: Write the first four terms of each of the following sequences whose general terms are given below: (i) a, = (–1)" 3" " 1 (ii) to - (n − 1)(n − 2)(n + 1) Solution: (i) General term is a, ...

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Saraswati Mathematics Term 1 2 1 tan” 0 = seco 0 – 1 = (*#) – 1 4x 1 1 1 1 2 2 = x* + —- + - – 1 = x* + - 16x” 2 16x” 2 2 1 = | x – — .. an o- (-k). 4x 4x cos” 0 – 1 + cos” 0 = seco ( – 1 1. -> 2 * 0 = 2 -> 2 = COS seco () cos” () 2 cos” 0 () 1 => COS () = –F– N2 ...

Author: Rajesh K. Dewan

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Saraswati Mathematics Vol 2 t 2 = a + (2–1) d = a + d and t3 = a +2d ∴ The product of 2nd and 3rd term = t2.t3 = (a + d )(a + 2d). Also the product of Ist and 4th term = t1.t4 = a(a + 3d) Hence, t2.t3 – t1.t4 = (a + d )(a + 2d) – a(a + 3d) = a2 + 3ad + 2d 2 ...

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Saraswati Mathematics Term 2 A Text book on maths

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Saraswati Mathematics (b) Find the H.C.F. of the three terms. (i) If the H.C.F. ≠ 1, then take out the H.C.F. as common factor. (ii) If the H.C.F. = 1, then two terms i.e., the first and the third term must be perfect squares proceded by the same sign i.e., ...

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Saraswati Mathematics Vol 1 These triangles formed using above patterns are called Pascal's triangles. 2. The first term in the (Binomial) expansion [i.e., the first term of the (binomial)index] is xn or nC0xn y0. 3. The second term is nxn – 1 y1 =nC1 xn– 1 y1. 4.

Author: O P Chugh, N P Bali

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Me n Mine Mathematics Term 2 Yes/No 2. Is the work done by the student precise, neat and accurate? Yes/No 3. Is the student able to interpret word problems into mathematical form? Yes/No 4. Is the student capable of interpreting data? Yes/No 5.

Author: Saraswati Experts

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Me n Mine Mathematics Term 2 TM Saraswati Experts. 10. 11. 12. 13. 14. Solutions to Value Based Questions 45, 45 – x, 45. an = x a + (n − 1) x d = x 1 + (n − 1) x 3 = x 311 – 2 = x ...(i) o - #(a+1) => 590 = #(1+x) 1180 = n(1 + 3n - 2) 1180 = 3n2 – n E} 3n2 – n ...

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Me n Mine Mathematics Term 1 Saraswati Experts. Adding (i) and (ii), we get x * * = J5 + 2 + N5 - 2 = 2 J5 3C Squaring both sides, we get 1\? (, .) = (2 J5)* => * * : 2, 1-4 & 5 => *** 12–20 3C % % 3C 2, 1 EX x + → = 20–2 = 18 % Thus, x” + * = 18. Hence proved.

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Reflections Termbook Class 03 Term 02 Reflections An Integrated Term Course General Knowledge Social Studies Mathematics 3 Term 2 English Science Rita Wilson ... MA (Mathematics), BEd Anju Loomba, MA (Mathematics), BEd Ashoo Kalra, BA, BEd (An imprint of New Saraswati House ...

Author: Rita Wilson, Milan Gowel, Kusum Wadhwa, Anju Loomba

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