Mathematics-6.Unit-07-Progression & Series-Test With Soultion

1. ASSIGNMENT SOLUTION 1. (A) The rth term of the series is given by Tr = (n –r + 1)r Sum of the series = ⇒ Sn = ⇒ Sn= = 2. (B) 3. (A) a1, a2 , a3 , . .. . . . , an are in H. P. Then are in A.P. ⇒ = d (say) . . . . (1) ⇒ a1a2 + a2a3 + a3a4 + . . . . + an-1an = (a1 -an) . .. . . (2) If we add all (n-1) terms of (1), we get d ⇒ . Thus from (2) a1a2 + a2a3 + a3a4 +. . .. . . . + an-1an = ( n-1)a1an . 4. (C) 5. (B) Sn = 5n2 + 2n, Sn –1 = 5(n –1)2 + 2(n –1) ⇒ Tn = Sn –Sn –1 = 10n –3 ⇒ T2 = 20 –3 = 17 6. (B) 2 7. (B) Let Sn and be sum of n terms of two series then ⇒ a1 = 1, d1 = 1, a2 = 2, d2 = 1 ⇒ . 8. (C) Let b = ar, c = a For triangle ABC, a + b > c a + ar > a - r – 1 < 0 < r < Also, b + c > a a (r + ) > a + r – 1 > 0 r Finally, a + c > b a [1 + ) > ar - r + 1 > 0. which is true for all values of r. In this case ‘r’ can not be negative r 9. (C) Given that a, b, c are in G.P ⇒ b2 = ac ⇒ 2 logx b = logx a + logx c ⇒ ⇒ loga x, logb x, logc x are in H.P. 10. (A) Using A.M. H.M. we get > b, a + c > 2b and > c, b + d > 2c a + c + b + d > 2b + 2c a + d > b + c 11. (B) The series can be written as S = (1 + 3 + 5 + 7 + …..n/2 terms) + (4 + 6 + 8 + 10 + …..n/2 terms) S = = [n] + [8 + n –2] = = 12. (B, D) It is easy to see that Δ = (b2 –ac)(aα2 +2bα +c) . Hence Δ =0 if a, b, c are in G.P. or α is a root of ax2 +2bx +c = 0. 13. (A) If n is odd, n-1 is even. Sum of (n-1) terms will be . The nth term will be n2. Hence the required sum = +n2 = 14. (B) = 6 + (4d + 9d + 14d + 19d + 23d) = 3 (2 + 23d) = 225 (given) Now, (2 + 23d) = 12 75 = 900 15. (A) The rth A. P. has first term r and common difference 2r – 1. Hence sum of its n terms = . The required sum = = = = . 16. (C) = = = (say) ( .5 + 5 + 9) = 15 Clearly, is the point of minima for f( ) d = 17. (B) Let the first five terms of the given G.P. be a1 , a2 , a3 , a4, a5 . Hence a3 = 4 . Now a1 a5 = a2 a4 = a32 ⇒ a1 a2 a3 a4 a5 = 45. 18. (D) Let the numbers be a1, a2, a3, . . ., an. Then a1.a2.a3 . .….an =1. Using A.M. ≥ G.M , we get ⇒ a1 + a2 + a3 + . . . . + an ≥ n 19. (C) a, b, c are in H.P. ⇒ b = ⇒ ⇒ . . . . (A) Again a, b, c are in H.P. ⇒ b = ⇒ ⇒ . . . . (B) From (A) and (B) = 2. 20. (C, D) Let α be the first and β be the (2n-1) terns of an A.P. , G.P. and H.P. , then α, a, β will be in A.P. , α, b, β will be G.P. α, c, β will be in H.P. Hence a, b, c are respectively A. M. , G.M. and H.M. of  and . Since A.M. ≥ G.M. ≥H.M. , a ≥ b ≥ c. Again a = , b2 = αβ and c = . Hence ac-b2 =0. 21. (A) Using A. M. ≥ G. M. ⇒ ≥ 3. 22. (D) 2b = a + c 8 = = + + 3ac (a + c) 8 = + + 3ac(2b) + - 8 = -6 abc 23. (A) S = … ∞ = = 24. (B) 25/26. Sn = = = 6 = Ans. of 25 question Now, S∞ = Sn = = = 6 Ans. of 26 question 27. (A) ⇒ − 2 = 0 28. (A) a = a1 + (p − 1) d … (1) b = a1 + (q − 1) d … (2) where a1 is the first term and d is the common difference of the A.P. Now, a − b = (p − q) d ⇒ d = and a + b = 2 a1 + (p + q − 2) d = 2 a1 + (p + q − 1) d − d … (3) Sp+q = [2 a1 + (p + q − 1) d] = [a + b + d] = 29. (C) Let ‘A’ be first term and ‘r’ be the common ratio. Then a = term 30. (C) S = + … ∞ = = 2. TEST SOLUTION 1. (B) Let the first term be ‘a’ and common ratio be ‘r’ 4 = a Required product = a. ar. a .a .a = . = 2. (B) 2 b = a + c ⇒ b − c = a − b … (1) ⇒ ⇒ {using equation (1)} ⇒ = ⇒ b ⇒ b (a + c) = − a c ⇒ b2 = { 2 b = a + c} ⇒ − , b, c are in G.P. 3. (A) = 4. (C) Since b is the H.M. of a and c, > b (A.M. > H.M.) Again c is the H.M. of b and d , > c ( A.M. > H.M.) Adding, we get + > b+c ⇒ a + d > b+ c. 5. (D) a + b + c = 18 2. + 3. + 4. = 18 Using weighted A.M. and G.M. inequality, we get 6. (B) 9 = 3 = 7. (C) 1 − , 1, 1 + are in A.P. ⇒ are in A.P. ⇒ are in A.P. 8. (A) b ac abc = 4 9. (B) Let S (n) be the sum of the integers divisible by n from 1 to 100 then S (2) = 2 + 4 + … + 100 = 2 {1 + 2 + 3 … + 50} = 50 × 51 = 2550 S (5) = 5 + 10 + … + 100 = 5 {1 + 2 + … + 20} = = 1050 S (10) = 10 + 20 + … + 100 = 10 {1 + 2 + … + 10} = = 550 Now S (2 ∪ 5) = S (2) + S (5) − S (2 ∩ 5) = S (2) + S (5) − S (10) = 2550 + 1050 − 550 = 3050 10. (B) (2x − 1)2 = 2. (2x + 3) ⇒ 22x − 4. 2x − 5 = 0 ⇒ (2x − 5) (2x + 1) = 0 ⇒ 2x + 1 ≠ 0 so 2x − 5 = 0 ⇒ x = log2 5 11. (C) 2b = a + c, c = 2bd = c(b + d) = (a + c)d bc + cd = ad + cd bc = ad 12. (B) ax = by = cz ⇒ x log a = y log b = z log c = k ⇒ x = , y = , z = now ⇒ 2 log b = log a + log c ⇒ b2 = ac 13. (B) (2a + (3n – 1)d) (2a + (n – 2)d) = (2a (3n – n + 1) + d (3n(3n – 1) – (n – 1)(n – 2)) = (2a (2n + 1) + d(8 – 2)) = a(2n + 1) + d (4 – 1) = (2n + 1) (a + (2n – 1)d) = a + (2n – 1)d = (2n + 1) 14. (A) Tr = Sn = = = = 15. (C) ⇒ 16. (A) Since a, b, c are in A.P., therefore are also in A.P. will also be in A.P. 17. (C) S = 2.357357357… = 2 + .357357 1000 S = 2357.357357 … = 2357 + .357357 … 1000 S = 2357 + S − 2 ⇒ 9995 = 2355 ⇒ S = 18. (B) If x, y, z are in GP then log x, log y, log z are in A.P. i.e. log x + 1, log y + 1, log z + 1 are in A.P. i.e. are in H.P. 19. (B) 20. (A) S = 5sinx−1 + 5−sinx−1 = now 5sinx + ≥ 2 ⇒ s ≥ 21. (D) S = 5 + 7 + 11 + 17 + 25 + …. + S = 5 + 7 + 11 + 17 + …. + + 0 = 5 + 2 + 4 + 6 + 8 …. - = 5 + 2[1 + 2 + 3 + 4 + …. + (n – 1) terms] = 5 + 2 n = n(n – 1) + 5 S = = ( - n + 5) = + 5n = (2 + 28) 22. (D) log4 x + log4 + log4 + … + log4 = log2 x ⇒ log4 x = log2 x ⇒ 1 + = 2 {Let n = 2k} ⇒ = 2 ⇒ = 0 ⇒ 2k = ∞ ⇒ n = ∞ 23. (C) are in A.P. ⇒ are in A.P. ⇒ are in A.P. ⇒ are in A.P. ⇒ are in A.P. ⇒ a, b, c are in H.P. 24. (B) ab + bc + cd = b(a + c) + cd = 2ac + cd = c(2a + d) 25. (A) Let S = 1 + 2 + 3 + 4 + 5 + … + 2n − 1 Now 1 + 2 + 3 + 4 + … 2n = = n (2n + 1) ⇒ S + 2 + 4 + 6 + … + 2n = n (2n + 1) ⇒ S + 2 {1 + 2 + 3 + … + n} = n (2n + 1) ⇒ S + n (n + 1) = n (2n + 1) ⇒ S = n2 26. (C) If 1 + a + a2 + … + an = (1 + a) (1 + a2) (1 + a4) then value of n will be the highest power of a, which will be 4 + 2 + 1 = 7 27. (C) T(r) = 1 - T(r) = = n - = n – 1 + 28. (C) S = … n terms = … n terms = (1 + 1 + 1 + … n terms) − = n − = n − 1 + = 2−n + n − 1 29. (D) S = 1 + 2 + … + n … (1) S = + 2 + … + (n − 1) + n … (2) Subtracting − + n − + n = n − n + n ⇒ S = n2 30. (A) S = + + + … 10 terms S = [1 + 31/2 + 3 + 33/2 + …… 10 terms] S = = 121 ( + )