Sequences and Series (Ex – 9.3) Question 1.Find the 20th and nth terms of the G.P. 5/2,5/4,5/8, …. Solution: Question 2.Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2. Solution:We have, as = 192, r = 2 Question 3.The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. […]

Sequences and Series (Ex – 9.2) Question 1.Find the sum of odd integers from 1 to 2001. Solution:We have to find 1 + 3 + 5 + ……….. + 2001This is an A.P. with first term a = 1, common difference d = 3-1 = 2 and last term l = 2001∴ l = a

Sequences and Series (Ex – 9.1) Question 1.an = n(n + 2) Solution:We haven an = n(n + 2)substituting n = 1, 2, 3, 4, 5, we geta1 = 9(1 + 2) = 1 x 3 = 3a2 = 2(2 + ) = 2 x 4 = 8a3 = 3(3 + 2) = 3 x 5 = 15a4 = 4(4 +

Binomial Theorem (Ex – 8.2) Question 1.Find the coefficient of x5 in (x + 3)8 Solution.Suppose x5 occurs in the (r + 1)th term of the expansion (x + 3)8 Question 2.a5 b7in (a-2b)12 Solution.Suppose a5 b7 occurs in the (r + 1)th term of the expansion (a – 2b)12. Write the general term in the expansion of Question 3.(x2 – y)6 Solution.

Binomial Theorem (Ex – 8.1) Expand each of the expressions in Exercises 1 to 5. Question 1.(1−2x)5 Solution. Question 2.(2/x−x/2)5 Solution. Question 3.(2x−3)6 Solution. Question 4.(x/3+1/x)5 Solution. Question 5.(x+1/x)6 Solution. Using binomial theorem, evaluate each of the following Question 6.(96)3 Solution. Question 7.(102)5 Solution. Question 8.(101)4 Solution. Question 9.(99)5 Solution. Question 10.Using Binomial Theorem, indicate

Permutations and Combinations (Ex – 7.4) Question 1.If nC8 = nC2, find nC2. Solution.We have, nC8 = nC2 Question 2.Determine n if(i) 2nC3: nC3 =12 : 1(ii) 2nC3: nC3= 11 : 1 Solution. Question 3.How many chords can be drawn through 21 points on a circle? Solution. Question 4.In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and

Permutations and Combinations (Ex – 7.3) Question 1.How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated? Solution.Total digits are 9. We have to form 3 digit numbers without repetition.∴ The required 3 digit numbers = 9P3=9!/6!=9×8×7×6!/6!=504 Question 2.How many 4-digit numbers are there with no digit

Permutations and Combinations (Ex – 7.2) Question 1.Evaluate(i) 8!(ii) 4!-3! Solution.(i) 8! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40320(ii) 4! – 3! = (1 x 2 x 3 x 4) – (1 x 2 x 3) = 24 – 6 = 18 Question 2.Is 3! +

Permutations and Combinations (Ex – 7.1) Question 1.How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that(i) repetition of the digits is allowed?(ii) repetition of the digits is not allowed? Solution.(i) (ii) When repetition is not allowed then first place can be filled in 5 different ways, second

Linear Inequalities (Ex – 6.3) Solve the following system of inequalities graphically: Question 1.x ≥ 3, y ≥ 2 Solution.x ≥ 3, y ≥ 2(i) AB represents the line x = 3Putting x = 0 in x ≥ 30 ≥ 3, which is not true.Therefore, origin does not lie in the region of x ≥ 3Its