Geometric Progression

 Problem : 1 [ Geometric Progression ]       Solve this type of problem 1. For given geometric progression series 3,6,12,24,48 ,... find 10 th term and addition of first 10 th terms. Show Answer A: Here a = 3,r = 6/3 = 2We know that, a_n = a × r^(n-1)a_10 = 3 × 2^(10 - 1)= 3 × 512= 1536We know that, S_n = a * (r^n - 1)/(r - 1):. S_10 = 3 × (2^10 - 1)/(2 - 1)=> S_10 = 3 × (1024 - 1)/1=> S_10 = 3 × 1023/1=> S_10 = 3069Hence, 10^(th) term of the given series is 1536 and sum of first 10^(th) term is 3069

 Problem : 2 [ Geometric Progression ]       Solve this type of problem 2. For given geometric progression series 3,6,12,24,48 ,... then find n such that S(n) = 3069 . Show Answer A: Here a = 3,r = 6/3 = 2We know that, S_n = a * (r^n - 1)/(r - 1)=> S_n = 3 × ((2)^n - 1) / (2 - 1)=> 3069 = 3 × ((2)^n - 1)/(1)=> 2^n - 1 = 3069 × (1) / 3=> 2^n - 1 = 1023=> 2^n = 1024=> 2^n = 2^10=> n = 10

 Problem : 3 [ Geometric Progression ]       Solve this type of problem 3. For given geometric progression series 3,6,12,24,48 ,... then find n such that f(n) = 1536 . Show Answer A: Here a = 3,r = 6/3 = 2Let n be the term such that f(n) = 1536We know that, a_n = a × r^(n-1)=> 3 × 2^(n-1) = 1536=> 2^(n-1) = 512=> 2^(n-1) = 2^9=> n - 1 = 9=> n = 9 + 1=> n = 10

 Problem : 4 [ Geometric Progression ]       Solve this type of problem 4. For geometric progression f( 1 ) = 2 , f( 4 ) = 54 then find f( 3 ) and S( 3 ). Show Answer A: We know that, a_n = a × r^(n-1)Here a_1 = 2=> a × r^(1 - 1) = 2=> a × r^0 = 2=> a = 2 ->(1)a_4 = 54=> a × r^(4 - 1) = 54=> a × r^3 = 54 ->(2)Solving (1) and (2), we get a = 2 and r = 3We know that, a_n = a × r^(n-1)a_3 = 2 × 3^(3 - 1)= 2 × 9= 18We know that, S_n = a * (r^n - 1)/(r - 1):. S_3 = 2 × (3^3 - 1)/(3 - 1)=> S_3 = 2 × (27 - 1)/2=> S_3 = 2 × 26/2=> S_3 = 26

 Problem : 5 [ Geometric Progression ]       Solve this type of problem 5. For geometric progression f( 1 ) = 2 , f( 4 ) = 54 , then find n such that f(n) = 18 . Show Answer A: We know that, a_n = a × r^(n-1)Here a_1 = 2=> a × r^(1 - 1) = 2=> a × r^0 = 2=> a = 2 ->(1)a_4 = 54=> a × r^(4 - 1) = 54=> a × r^3 = 54 ->(2)Solving (1) and (2), we get a = 2 and r = 3Let n be the term such that f(n) = 18We know that, a_n = a × r^(n-1)=> 2 × 3^(n-1) = 18=> 3^(n-1) = 9=> 3^(n-1) = 3^2=> n - 1 = 2=> n = 2 + 1=> n = 3

 Problem : 6 [ Geometric Progression ]       Solve this type of problem 6. For geometric progression addition of 3 terms is 26 and their multiplication is 216 , then that nos. Show Answer A: Let the terms are (a / r) , a , (a × r)Addition : (a / r) + a + (a × r) = 26 ->(1)Multiplication : (a / r) × a × (a × r) = 216=> a^3 = 216=> a = 6Now putting a = 6 in (1), we get(6 / r) + 6 + (6 × r) = 26(6 / r) + (6 × r) = 20=> r = 3 or r = 1/3Now, a = 6 and r = 3 => (6 / 3) , 6 , (6 × 3) => 2 , 6 , 18or a = 6 and r = 1/3 => (6 / (1/3)) , 6 , (6 × (1/3)) => 18 , 6 , 2

 Problem : 7 [ Geometric Progression ]       Solve this type of problem 7. For geometric progression multiplication of 5 terms is 1 and 5 th term is 81 times then the 1 th term. Show Answer A: Let the terms are (a / r^2) , (a / r) , a , (a × r) , (a × r^2)Multiplication : (a / r^2) * (a / r) * a * (a × r) * (a × r^2) = 1=> a^5 = 1=> a = 1Now 5^(th) term is 81 times than the 1^(st) term=> (a × r^2) = 81 (a / r^2)=> r^4 = 81=> r = 3Now, a = 1 and r = 3 => (1 / 3^2) , (1 / 3) , 1 , (1 × 3) , (1 × 3^2) => 0.1111 , 0.3333 , 1 , 3 , 9

 Problem : 8 [ Geometric Progression ]       Solve this type of problem 8. Arithmetic mean of two no is 13 and geometric mean is 12 , then find that nos. Show Answer A: Let the two terms be a and b.=> A = (a + b)/2 = 13 and G = sqrt(ab) = 12=> a + b = 26 ->(1) and ab = 144 ->(2)=> b = 26 - a ->(3)Now putting the value of (3) in (2), we get a(26 - a) = 144=> 26a - a^2 = 144=> a^2 - 26a + 144 = 0=> (a - 8)(a - 18) = 0=> a = 8 or a = 18=> a = 8 => b = 26 - a = 26 - 8 = 18and a = 18 => b = 26 - a = 26 - 18 = 8:. Required numbers are 8 and 18

 Problem : 9 [ Geometric Progression ]       Solve this type of problem 9. Two nos are in the ratio 9 : 16 and difference of arithmetic mean and geometric mean is 1 , then find that nos. Show Answer A: Here ratio of the two terms is 9:16.Let the two terms be 9x and 16x.Now difference of their Arithmetic mean A and Geometric mean G is 1.=> A - G = 1 (Because A > G).=> (9x + 16x)/2 - sqrt(9x × 16x) = 1.=> (25x)/2 - sqrt(144) x = 1.=> x = 2=> The required two terms are 9 × 2 and 16 × 2.=> The required two terms are 18 and 32.

 Problem : 10 [ Geometric Progression ]       Solve this type of problem 10. Find 6 arithmetic mean between 3 and 24 . Show Answer A: Let  A_1, A_2, A_3, A_4, A_5, A_6 be the 6 arithmetic mean between 3 and 24.:. 3, A_1, A_2, A_3, A_4, A_5, A_6, 24 are in AP .Here First term a = 3 and Last term b = 24Comman difference d = (b-a)/(n+1) = (24-3)/(6+1) = 21/7 = 3:. Required Arithemetic Means, A_1 = T_2 = a + d = 3 + 3 = 6A_2 = T_3 = a + 2d = 3 + 2(3) = 9A_3 = T_4 = a + 3d = 3 + 3(3) = 12A_4 = T_5 = a + 4d = 3 + 4(3) = 15A_5 = T_6 = a + 5d = 3 + 5(3) = 18A_6 = T_7 = a + 6d = 3 + 6(3) = 21

 Problem : 11 [ Geometric Progression ]       Solve this type of problem 11. Find 3 geometric mean between 1 and 256 . Show Answer A: Let  a_1, a_2, a_3 be the 3 geometric mean between 1 and 256.:. 1, a_1, a_2, a_3, 256 are in GP .Here First term a = 1 and Last term b = 256Comman ratio r = (b/a)^(1/(n+1)) = (256/1)^(1/4) = 4:. Required Geometic Means, a_1 = a * r = 1 * 4 = 4a_2 = a r^2 = 1 × 4^2 = 16a_3 = a r^3 = 1 × 4^3 = 64

 Problem : 12 [ Geometric Progression ]       Solve this type of problem 12. Prove that 1 + (1 + 2) + (1 + 2 + 3) + ... n terms = n/6 (n + 1) (n + 2) Show Answer A: L.H.S. = 1 + (1 + 2) + (1 + 2 + 3) + ... + n terms= sum [ f(n) ]= sum [ sum n ]= sum [ n/2 (n + 1) ]= 1/2 sum n^2 + 1/2 sum n= 1/2 * (n (n + 1) (2n + 1))/6 + 1/2 * (n (n + 1))/2= (n (n + 1) (2n + 1))/12 + (n (n + 1))/4= (n (n + 1) (2n + 1 + 3))/12= (n (n + 1) (2n + 4))/12= (n (n + 1) (n + 2))/6= R.H.S. (Proved)

 Problem : 13 [ Geometric Progression ]       Solve this type of problem 13. Prove that 1 × (22 - 32) + 2 × (32 - 42) + 3 × (42 - 52) + ... n terms = n/6 (n + 1) (4n + 11) Show Answer A: L.H.S. = 1 × (2^2 - 3^2) + 2 × (3^2 - 4^2) + 3 × (4^2 - 5^2) + ... n terms= sum [ f(n) ]= sum [ n ((n + 1)^2 - (n + 2)^2) ]= sum [ n (n^2 + 2n + 1 - n^2 - 2n - 4) ]= sum [ n (-2n - 3) ]= sum (-2n^2 - 3n) ]= -2 sum n^2 - 3 sum n= -2 * (n (n + 1) (2n + 1))/6 - 3 * (n (n + 1))/2= - n/6 (n + 1) [ 2(2n + 1) + 9 ]= - n/6 (n + 1) (4n + 11)= R.H.S. (Proved)

 Problem : 14 [ Geometric Progression ]       Solve this type of problem 14. 1 + x4 + 32 + 4  + x6 + 62 + 7  + x8 + 92 + ... 3n terms Show Answer A: S_(3n) = 1 + x^4 + 3^2 + 4 + x^6 + 6^2 + 7 + x^8 + 9^2 + ... 3n terms= (1 + 4 + 7 + ... n " terms") + (x^4 + x^6 + x^8 + ... n " terms") + (3^2 + 6^2 + 9^2+ ... n " terms")= (1 + 4 + 7 + ... n " terms") + x^4 (1 + x^2 + x^4 + ... n " terms") + 3^2 (1 + 2^2 + 3^2 + ... n " terms")= n/2 [ 2 × 1 + (n - 1) × 3 ] + x^4 [ (x^2)^n - 1 ] / ( x^2 - 1) + 9 × n/6 (n + 1) (2n + 1)= n/2 (3n - 1) + x^4 (x^(2n) - 1) / ( x^2 - 1) + (3n)/2 (n + 1) (2n + 1)

 Problem : 15 [ Geometric Progression ]       Solve this type of problem 15. 1 + (1 + 3) + (1 + 3 + 5) + ... n terms Show Answer A: S_n = sum [ f(n) ]= sum [ sum (2n - 1) ]= sum [ 2 sum n - sum 1 ]= sum [ 2 * (n (n + 1))/2 - n ]= sum [ n^2 + n - n ]= sum n^2= (n (n + 1) (2n + 1))/6

 Problem : 16 [ Geometric Progression ]       Solve this type of problem 16. Prove that for all n belongs to N, 12 × n + 22 (n - 1) + 32 (n - 2) + ... + n2 × 1 = n/12 (n + 1)2 (n + 2) Show Answer A: Here, f(r) = r^2 (n -r + 1) = (n + 1) r^2 - r^3Now, L.H.S. = 1^2 × n + 2^2 (n - 1) + 3^2 (n - 2) + ... + n^2 × 1= sum [ (n + 1) r^2 - r^3 ] (Where r = 1 to n)= (n + 1) sum r^2 - sum r^3 (Where r = 1 to n)= (n + 1) * (n (n + 1) (2n + 1))/6 - (n^2 (n + 1)^2)/4= (n (n + 1)^2 [ 2 (2n + 1) - 3 n])/12= (n (n + 1)^2 (n + 2))/12= R.H.S. (Proved)

 Problem : 17 [ Geometric Progression ]       Solve this type of problem 17. Prove that 1 × 22 + 3 × 52 + 5 × 82 + ... n terms = n/2 (9n3 + 4n2 - 4n -1) Show Answer A: L.H.S. = 1 × 2^2 + 3 × 5^2 + 5 × 8^2 + ... n terms= sum [ f(n) ]= sum [ (2n - 1)(3n - 1)^2 ]= sum [ (2n - 1)(9n^2 - 6n + 1)]= sum [ 18n^3 - 21n^2 + 8n - 1]= 18 sum n^3 - 21 sum n^2 + 8 sum n - sum 1= 18 * (n^2 (n+1)^2)/4 - 21 * (n (n + 1) (2n + 1))/6 + 8 * (n (n+1))/2 - n= n/2 [ 9 n (n + 1)^2 - 7(n + 1) (2n + 1) + 8 (n+1) - 2 ]= n/2 [ 9 n (n^2 + 2n +1) - 7 (2n^2 + 3n + 1) + 8 n + 8 - 2 ]= n/2 [ 9 n^3 + 18 n^2 + 9n - 14n^2 - 21n - 7 + 8 n + 8 - 2 ]= n/2 [ 9 n^3 + 4 n^2 - 4n - 1 ]= R.H.S. (Proved)

 Problem : 18 [ Geometric Progression ]       Solve this type of problem 18. Prove that 12 + (12 + 22) + (12 + 22 + 32) + ... n terms = n/12 (n + 1)2 (n + 2) Show Answer A: L.H.S. = 1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ... n terms= sum [ f(n) ]= sum [ sum n^2 ]= sum [ n/6 (n + 1) (2n + 1) ]= sum [ n/6 (2n^2 + 3n + 1) ]= sum [ 1/6 (2n^3 + 3n^2 + n) ]= 1/6 [ 2 sum n^3 + 3 sum n^2 + sum n ]= 1/6 [ 2 * (n^2 (n+1)^2)/4 + 3 * (n (n + 1) (2n + 1))/6 + (n (n + 1))/2 ]= 1/6 [ (n^2 (n+1)^2)/2 + (n (n + 1) (2n + 1))/2 + (n (n + 1))/2 ]= 1/6 * (n (n + 1))/2 [ n (n+1) + (2n + 1) + 1 ]= n/12 (n + 1) [ n^2 + n + 2n + 2 ]= n/12 (n + 1) [ n^2 + 3n + 2 ]= n/12 (n + 1) [ (n + 2) (n + 1) ]= n/12 (n + 1)^2 (n + 2)= R.H.S. (Proved)

 Problem : 19 [ Geometric Progression ]       Solve this type of problem 19. Prove that 2 + 5 + 10 + 17 + ... n terms =n/6 (2n2 + 3n + 7) Show Answer A: L.H.S. = 2 + 5 + 10 + 17 + ... n terms= sum [ 1 + n^2 ]= sum 1 + sum n^2= n + (n (n + 1) (2n + 1))/6= n/6 [ 6 + (2n^2 + 3n + 1) ]= n/6 (2n^2 + 3n + 7)= R.H.S. (Proved)

 Problem : 20 [ Geometric Progression ]       Solve this type of problem 20. Prove that S [ S (2n -3) ] =n/6 (n + 1)(2n - 5) Show Answer A: L.H.S. = sum [ sum (2n -3) ]= sum [ sum 2n - sum 3 ]= sum [ 2 × n/2 (n + 1) -3n ]= sum [ n (n + 1) - 3n ]= sum [ n^2 + n - 3n ]= sum [ n^2 - 2n ]= sum n^2 - 2 sum n= (n (n + 1) (2n + 1))/6 - 2 * (n (n + 1))/2= (n (n + 1) [ (2n + 1) - 6 ])/6= (n (n + 1) (2n - 5))/6= R.H.S. (Proved)

 Problem : 21 [ Geometric Progression ]       Solve this type of problem 21. For geometric progression, find 1 + 1/sqrt(2) + 1/2 + 1/2*sqrt(2) + ... 10 terms ( 21. For geometric progression, find 1 + 1/sqrt(x) + 1/x + 1/x*sqrt(x) + ... n terms where x = 2 and n = 10 . ) Show Answer A: Here a = 1, r = 1 / sqrt(2), n = 10We know that, S_n = a * (1 - r^n)/(1 - r):. S_10 = 1 × (1 - (1 / sqrt(2))^10) / (1 - 1 / sqrt(2))= 1 × (1 - (1 / 2)^5) / ((sqrt(2) - 1) / sqrt(2))= sqrt(2) × 1 × (1 - 1 / 32) / (sqrt(2) - 1) = sqrt(2) × 1 × ((32 - 1) / 32) / (sqrt(2) - 1) × (sqrt(2) + 1) / (sqrt(2) + 1)= sqrt(2) × 1 × (31/32) / (2 - 1) × (sqrt(2) + 1)= (31/32) × sqrt(2) × (sqrt(2) + 1)= (31/32) × (2 + sqrt(2))

 Problem : 22 [ Geometric Progression ]       Solve this type of problem 22. Find 1^2 + 2^2 + ...+ 10^2 , ( 22. Find a^2 + b^2 + ...+ n^2 , where a = 1 , b = 2 and n = 10 . ) Show Answer A: Here Find  1^2 + 2^2 + 3^2 + ... + 10^2 = sum i^2 (where i = 1,...,10)= (n (n + 1)(2n + 1)) / 6 (where n = 10) = (10 (10 + 1) (2×10 + 1)) / 6 = (10 × 11 × 21) / 6= 5 × 11 × 7= 385

 Problem : 23 [ Geometric Progression ]       Solve this type of problem 23. Find 1^2 + 2^2 + ... 10 terms, ( 23. Find a^2 + b^2 + ... n terms, where a = 1 , b = 2 and n = 10 . ) Show Answer A: Here Find  1^2 + 2^2 + 3^2 + ... 10 "terms" = sum i^2 (where i = 1,...,10)= (n (n + 1)(2n + 1)) / 6 (where n = 10) = (10 (10 + 1) (2×10 + 1)) / 6 = (10 × 11 × 21) / 6= 5 × 11 × 7= 385