Q: For the functions f( ) = ^2 + ^2 , and g( ) = ^2 + ^2 , > > 0, let _ 0 < < /2 f( ) = _ 0 < < g( ). If the first term of a G.P. is ( 2 ), its common ratio is ( 2 ) and the sum of its first 10 terms is

For the function f( ) = ^2 + ^2 , applying the AM-GM inequality gives: ^2 + ^2 2 ^2 ^2 = 2 Thus, _ 0 For the function g( ) = ^2 + ^2 , we can rewrite it as: g( ) = ^2 + (1 - ^2 ) = + ( - ) ^2 Since > > 0, the maximum value of g( ) occurs when ^2 is maximum, which is 1 (at = /2). Thus, _ 0 Given that f( ) = g( ), we have: 2 = Squaring both sides (since > 0), we get: 4 = ^2 = 4 For the given Geometric Progression (G.P.): First term a = 2 = 4 2 = 2 Common ratio r = 2 = 2 4 = 1 2 The sum of the firs

Question 1

Let = 3+4+8+9+13+14+ upto 40 terms. If ( )^ 1020 is a root of the equation x^2+x-2=0, (0, 2 ), then ^2 + 3 ^2 is equal to:

Accepted Answer

The given series is = 3+4+8+9+13+14+ up to 40 terms. We can split the series into two arithmetic progressions, each containing 20 terms: S_1 = 3 + 8 + 13 + up to 20 terms S_2 = 4 + 9 + 14 + up to 20 terms Using the sum formula for an AP, S_n = n 2 [2a + (n-1)d]: S_1 = 20 2 [2(3) + (20-1)5] = 10[6 + 95] = 1010 S_2 = 20 2 [2(4) + (20-1)5] = 10[8 + 95] = 1030 Therefore, = S_1 + S_2 = 1010 + 1030 = 2040. The exponent in the given expression is 1020 = 2040 1020 = 2. Thus, ( )^2 = ^2 is a root of the

Question 2

The sum 1 + 1 2 (1^2 + 2^2) + 1 3 (1^2 + 2^2 + 3^2) + upto 10 terms is equal to :

Accepted Answer

The n-th term of the given series is: T_n = 1 n (1^2 + 2^2 + + n^2) Using the formula for the sum of squares of the first n natural numbers: T_n = 1 n ( n(n+1)(2n+1) 6 ) = (n+1)(2n+1) 6 = 2n^2 + 3n + 1 6 The sum of the first 10 terms is S_ 10 = _ n=1 ^ 10 T_n S_ 10 = 1 6 ( 2 _ n=1 ^ 10 n^2 + 3 _ n=1 ^ 10 n + _ n=1 ^ 10 1 ) Using the standard summation formulas: _ n=1 ^ 10 n^2 = 10 11 21 6 = 385 _ n=1 ^ 10 n = 10 11 2 = 55 _ n=1 ^ 10 1 = 10 Substituting these values into the sum expression: S_ 10

Question 3

The value of 1^3 - 2^3 + 3^3 - + 15^3 is:

Accepted Answer

The given series is S = 1^3 - 2^3 + 3^3 - 4^3 + + 15^3. This can be rewritten as the sum of all cubes up to 15 minus twice the sum of cubes of even numbers up to 14: S = (1^3 + 2^3 + 3^3 + + 15^3) - 2(2^3 + 4^3 + 6^3 + + 14^3) S = _ k=1 ^ 15 k^3 - 2 2^3 (1^3 + 2^3 + 3^3 + + 7^3) S = _ k=1 ^ 15 k^3 - 16 _ k=1 ^ 7 k^3 Using the formula for the sum of cubes _ k=1 ^ n k^3 = ( n(n+1) 2 )^2: _ k=1 ^ 15 k^3 = ( 15 16 2 )^2 = (120)^2 = 14400 _ k=1 ^ 7 k^3 = ( 7 8 2 )^2 = (28)^2 = 784 Substituting these

Question 4

The sum of the first ten terms of an A.P. is 160 and the sum of the first two terms of a G.P. is 8. If the first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P

Accepted Answer

Let the first term of the A.P. be a and its common difference be d. Let the first term of the G.P. be A and its common ratio be R. Given that the sum of the first ten terms of the A.P. is 160: 10 2 [2a + (10 - 1)d] = 160 5(2a + 9d) = 160 2a + 9d = 32 Given that the sum of the first two terms of the G.P. is 8: A + AR = 8 A(1 + R) = 8 We are also given that a = R and A = d. Substituting these into the first equation: 2R + 9A = 32 R = 32 - 9A 2 Substitute R into the second equation: A (1 + 32 - 9A

Question 5

For the functions f( ) = ^2 + ^2 , and g( ) = ^2 + ^2 , > > 0, let _ 0 < < /2 f( ) = _ 0 < < g( ). If the first term of a G.P. is ( 2 ), its common ratio is ( 2 ) and the sum of its first 10 terms is

Accepted Answer

For the function f( ) = ^2 + ^2 , applying the AM-GM inequality gives: ^2 + ^2 2 ^2 ^2 = 2 Thus, _ 0 For the function g( ) = ^2 + ^2 , we can rewrite it as: g( ) = ^2 + (1 - ^2 ) = + ( - ) ^2 Since > > 0, the maximum value of g( ) occurs when ^2 is maximum, which is 1 (at = /2). Thus, _ 0 Given that f( ) = g( ), we have: 2 = Squaring both sides (since > 0), we get: 4 = ^2 = 4 For the given Geometric Progression (G.P.): First term a = 2 = 4 2 = 2 Common ratio r = 2 = 2 4 = 1 2 The sum of the firs

Sequences and Series — Previous Year Questions & Solutions