Q: Let R denote the set of all real numbers. Let f: R R be a function such that f(x)>0 for all x R , and f(x+y)=f(x) f(y) for all x, y R . Let the real numbers a _1, a _2 , a _ 50 be in an arithmetic pro

aligned & f(x+y)=f(x) f(y) \\ & f(x)=k^x (f(x)>0 x R) \\ & f (a_ 31 )=64 f (a_ 25 ) aligned aligned & k ^ ( a +30 ~d ) =64 k ^ ( a +24 ~d ) \\ & k ^ 6 ~d =64 \\ & k ^ d =2 aligned _ i =1 ^ 50 f ( a _ i )= f ( a _1 )+ f ( a _2 )+ + f ( a _ 50 ) aligned & = k ^ a + k ^ a + d + + k ^ a +49 ~d = k ^ a ( k ^ 50 ~d -1 ) k ^ d -1 \\ & = k ^ a (2^ 50 -1 )=3 (2^ 25 +1 )( Given ) \\ & k ^ a = 3 2^ 25 -1 aligned aligned & _ i =6 ^ 30 f ( a _ i )= k ^ a +5 ~d + k ^ a +6 ~d + + k ^ a +29 ~d \\ & = k ^ a +5 ~

Q: Let 7 5 … 5 ⏞ 7 r denote the r + 2 digit number where the first and the last digits are 7 and the remaining r digits are 5 . Consider the sum S = 77 + 757 + 7557 + … + 7 5 … 5

Given, S = 77 + 757 + 7557 + … + 7 5 … 5 ⏞ 7 98 ⇒ S = 7 × 10 + 7 + 7 × 100 + 5 × 10 + 7 + 7 × 1000 + 5 × 100 + 5 × 10 + 7 … + 7 5 … 5 ⏞ 7 98 ⇒ S = 7 10 + 10 2 + … + 10 99 + 50 1 + 11 + … + 111 … 1 ⏞ 98 + 7 × 99 ⇒ S = 70 10 99 - 1 9 + 50 9 10 - 1 + 10 2 - 1 + … + 10 98 - 1 + 7 × 99 ⇒ S = 70 10 99 - 1 9 + 50 9 10 10 98 - 1 9 - 98 + 7 × 99 ⇒ S = 7 × 10 10

Q: Paragraph: Let M= \ (x, y) R R : x^ 2 +y^ 2 r^ 2 \ , where r>0 . Consider the geometric progression a_ n = 1 2^ n-1 , n=1,2,3, . Let S_ 0 =0 and, for n 1, let S_ n denote the sum of the first n terms

S n = 1 + 1 2 + 1 2 2 + … + 1 2 n - 1 = 2 1 - 1 2 n = 2 - 1 2 n - 1 Centre of C n is 2 - 1 2 n - 2 , 0 and radius of C n is 1 2 n - 1 when r = 1025 513 < 2 C n will lie inside m when 2 - 1 2 n - 2 + 1 2 n - 1 < 1025 513 ⇒ 1 - 1 2 n < 1025 1026 ⇒ 2 n < 1026 ⇒ n ≤ 10 Hence number of circles ⇒ k = 10 Also ℓ = 5 3 k + 2 ℓ = 30 + 10 = 40

Q: Paragraph: Let M= \ (x, y) R R : x^ 2 +y^ 2 r^ 2 \ , where r>0 . Consider the geometric progression a_ n = 1 2^ n-1 , n=1,2,3, . Let S_ 0 =0 and, for n 1, let S_ n denote the sum of the first n terms

2 S n - 1 + a n < 2 199 - 1 2 198 2 2 2 - 1 2 n - 2 + 1 2 n - 1 < 2 199 - 1 2 198 2 2 2 - 1 2 n - 2 2 + 1 2 n - 1 < 2 2 - 2 2 198 1 2 n - 2 1 2 - 2 < - 2 2 198 1 2 n - 2 ( 2 2 - 1 ) 2 > 2 2 198 2 n - 2 < 2 - 1 2 2 197 n - 2 ≤ 197 number of circle = 199

Q: Let m be the minimum possible value of log 3 3 y 1 + 3 y 2 + 3 y 3 , where y 1 , y 2 , y 3 are real numbers for which y 1 + y 2 + y 3 = 9 . Let M be the maximum possible value of log 3 x

Applying A.M-G.M inequality, 3 y 1 + 3 y 2 + 3 y 3 3 ≥ 3 y 1 ⋅ 3 y 2 ⋅ 3 y 3 1 3 = 3 y 1 + y 2 + y 3 1 3 ⇒ 3 y 1 + 3 y 2 + 3 y 3 ≥ 81 So, m = log 3 81 = 4 Now, log 3 x 1 + log 3 x 2 + log 3 x 3 = log 3 x 1 . x 2 . x 3 ∵ x 1 + x 2 + x 3 3 ≥ x 1 . x 2 . x 3 1 3 ⇒ x 1 . x 2 . x 3 ≤ 27 (applying A.M-G.M inequality) So, M = log 3 27 = 3 Thus, log 2 m 3 + log 3 M 2 = 8

Q: Let a 1 , a 2 , a 3 , . . . . . . . . . be a sequence of positive integers in arithmetic progression with common difference 2 . Also, let b 1 , b 2 , b 3 ,

2 a 1 + a 2 + . . . . . . . . + a n = b 1 + b 2 + . . . . . . . . + b n ⇒ 2 n 2 2 a 1 + n − 1 ⋅ 2 = b 1 ⋅ 2 n − 1 2 − 1 ⇒ n 2 c + 2 n − 2 = c 2 n − 1 ⇒ 2 n c + n − 1 = c 2 n − 1 ⇒ c 2 n − 2 n − 1 = 2 n 2 − 2 n ⇒ c = 2 n 2 − 2 n 2 n − 2 n − 1 ≥ 1 . . . . . . . . . . . . . i ⇒ 2 n 2 − 2 n ≥ 2 n − 2 n − 1 ⇒ 2 n 2 + 1 ≥

Q: Let A P a ; d denote the set of all the terms of an infinite arithmetic progression with first term a and common difference d > 0 . If A P 1 ; 3 ∩ A P 2 ; 5 ∩ A P 3 ; 7 = A P a

Let T 1 , m : m t h term of 1 s t series A 1 ; 3 T 2 , n : n t h term of 2 n d series A 2 ; 5 T 3 , r : r t h term of 3 r d series A ( 3 ; 7 ) For common terms T ( 1 , m ) = T ( 2 , n ) = T ( 3 , r ) 1 + 3 m - 1 = 2 + 5 n - 1 = 3 + 7 r - 1 3 m - 2 = 5 n - 3 = 7 r - 4 ⇒ m = 5 n - 1 3 a n d r = 5 n + 1 7 For ‘ m ’ to be a natural number n = 2,5 , 11 , … And for ‘ r ’ to be a na

Question 1

Let R denote the set of all real numbers. Let f: R R be a function such that f(x)>0 for all x R , and f(x+y)=f(x) f(y) for all x, y R . Let the real numbers a _1, a _2 , a _ 50 be in an arithmetic pro

Accepted Answer

aligned & f(x+y)=f(x) f(y) \ & f(x)=k^x (f(x)>0 x R) \ & f (a_ 31 )=64 f (a_ 25 ) aligned aligned & k ^ ( a +30 ~d ) =64 k ^ ( a +24 ~d ) \ & k ^ 6 ~d =64 \ & k ^ d =2 aligned _ i =1 ^ 50 f ( a _ i )= f ( a _1 )+ f ( a _2 )+ + f ( a _ 50 ) aligned & = k ^ a + k ^ a + d + + k ^ a +49 ~d = k ^ a ( k ^ 50 ~d -1 ) k ^ d -1 \ & = k ^ a (2^ 50 -1 )=3 (2^ 25 +1 )( Given ) \ & k ^ a = 3 2^ 25 -1 aligned aligned & _ i =6 ^ 30 f ( a _ i )= k ^ a +5 ~d + k ^ a +6 ~d + + k ^ a +29 ~d \ & = k ^ a +5 ~

Question 2

Let 7 5 &#8230; 5 &#9182; 7 r denote the r + 2 digit number where the first and the last digits are 7 and the remaining r digits are 5 . Consider the sum S = 77 + 757 + 7557 + &#8230; + 7 5 &#8230; 5

Accepted Answer

Given, S = 77 + 757 + 7557 + &#8230; + 7 5 &#8230; 5 &#9182; 7 98 &#8658; S = 7 &#215; 10 + 7 + 7 &#215; 100 + 5 &#215; 10 + 7 + 7 &#215; 1000 + 5 &#215; 100 + 5 &#215; 10 + 7 &#8230; + 7 5 &#8230; 5 &#9182; 7 98 &#8658; S = 7 10 + 10 2 + &#8230; + 10 99 + 50 1 + 11 + &#8230; + 111 &#8230; 1 &#9182; 98 + 7 &#215; 99 &#8658; S = 70 10 99 - 1 9 + 50 9 10 - 1 + 10 2 - 1 + &#8230; + 10 98 - 1 + 7 &#215; 99 &#8658; S = 70 10 99 - 1 9 + 50 9 10 10 98 - 1 9 - 98 + 7 &#215; 99 &#8658; S = 7 &#215; 10 10

Question 3

Let l 1 , l 2 . . . l 100  be consecutive terms of an arithmetic progression with common difference d 1 , and let w 1 , w 2 , &#8230; , w 100  be consecutive terms of another arithmetic prog

Accepted Answer

Given  l 1 , l 2 . . . l 100  are consecutive terms of an  A . P Now let  T 1 = a  and common difference  = d 1   And similarly for A . P   w 1 , w 2 , . . . w 100 ,  T 1 = b  and common difference  = d 2 Now given,  A 51 - A 50 = l 51 w 51 - l 50 w 50 &#8658; a + 50 d 1 b + 50 d 2 - a + 49 d 1 b + 49 d 2 = 1000 &#8658; 50 b d 1 + 50 a d 2 + 2500 d 1 d 2 - 49 a d 2 - 49 b d 1 - 2401 d 1 d 2 = 1000 &#8658; b d 1 + a d 2 + 99 d 1 d 2

Question 4

Let a 1 , a 2 , a 3 , &#8230;  be an arithmetic progression with a 1 = 7  and common difference 8 . Let T 1 , T 2 , T 3 , &#8230;  be such that T 1 = 3  and  T n + 1 - T n = a

Accepted Answer

Given, a n = 7 + n - 1 8  and  T 1 = 3 Also  T n + 1 = T n + a n Now  T n = T n - 1 + a n - 1 Putting  n = 1  we get,  T 2 = T 1 + a 1 Now putting  n = 2  we get, T 3 = T 2 + a 2 &#8658; T 3 = T 1 + a 1 + a 2 And so on we get, &#8658; &#160; T n + 1 = T 1 + a 1 + a 2 + &#8230; + a n &#8658; &#160; T n + 1 = T 1 + n 2 2 7 + n - 1 8 &#8658; &#160; T n + 1 = T 1 + n 4 n + 3 &#160; &#160; &#160; &#160; &#160; &#8943; 1 Now for  n = 19  we get, T

Question 5

Paragraph: Let M= \ (x, y) R R : x^ 2 +y^ 2 r^ 2 \ , where r>0 . Consider the geometric progression a_ n = 1 2^ n-1 , n=1,2,3, . Let S_ 0 =0 and, for n 1, let S_ n denote the sum of the first n terms

Accepted Answer

S n = 1 + 1 2 + 1 2 2 + &#8230; + 1 2 n - 1 = 2 1 - 1 2 n = 2 - 1 2 n - 1 Centre of C n is 2 - 1 2 n - 2 , 0 and radius of C n is 1 2 n - 1 when r = 1025 &#160; 513 &#60; 2 C n will lie inside  m when 2 - 1 2 n - 2 + 1 2 n - 1 &#60; 1025 &#160; 513 &#8658; 1 - 1 2 n &#60; 1025 1026 &#8658; 2 n &#60; 1026 &#8658; n &#8804; 10 Hence number of circles &#8658; k = 10 Also &#8467; = 5 3 k + 2 &#8467; = 30 + 10 = 40

Question 6

Paragraph: Let M= \ (x, y) R R : x^ 2 +y^ 2 r^ 2 \ , where r>0 . Consider the geometric progression a_ n = 1 2^ n-1 , n=1,2,3, . Let S_ 0 =0 and, for n 1, let S_ n denote the sum of the first n terms

Accepted Answer

2 &#160; S n - 1 + a n &#60; 2 199 - 1 2 198 2 2 2 - 1 2 n - 2 + 1 2 n - 1 &#60; 2 199 - 1 2 198 2 2 2 - 1 2 n - 2 2 + 1 2 n - 1 &#60; 2 2 - 2 2 198 1 2 n - 2 1 2 - 2 &#60; - 2 2 198 1 2 n - 2 ( 2 2 - 1 ) 2 &#62; 2 2 198 2 n - 2 &#60; 2 - 1 2 2 197 n - 2 &#8804; 197 number of circle &#160; = 199

Question 7

Let m be the minimum possible value of log 3 3 y 1 + 3 y 2 + 3 y 3 , where y 1 , &#160; y 2 , &#160; y 3 are real numbers for which y 1 + y 2 + y 3 = 9 . Let M be the maximum possible value of log 3 x

Accepted Answer

Applying A.M-G.M inequality, 3 y 1 + 3 y 2 + 3 y 3 3 &#8805; 3 y 1 &#8901; 3 y 2 &#8901; 3 y 3 1 3 = 3 y 1 + y 2 + y 3 1 3 &#8658; 3 y 1 + 3 y 2 + 3 y 3 &#8805; 81 So,  m = log 3 81 = 4 Now,  log 3 x 1 + log 3 x 2 + log 3 x 3 = log 3 x 1 . x 2 . &#160; x 3 &#8757; x 1 + x 2 + x 3 3 &#8805; x 1 . x 2 . x 3 1 3 &#8658; x 1 . x 2 . x 3 &#8804; 27  (applying A.M-G.M inequality) So,  M = log 3 27 = 3 Thus, log 2 m 3 + log 3 M 2 = 8

Question 8

Let a 1 , &#160; a 2 , &#160; a 3 , &#160; . . . . . . . . .  be a sequence of positive integers in arithmetic progression with common difference 2 . Also, let b 1 , &#160; b 2 , &#160; b 3 ,

Accepted Answer

2 &#160; a 1 + a 2 + . . . . . . . . + a n = b 1 + b 2 + . . . . . . . . + b n &#8658; 2 n 2 2 a 1 + n &#8722; 1 &#8901; 2 = b 1 &#8901; 2 n &#8722; 1 2 &#8722; 1 &#8658; n 2 c + 2 n &#8722; 2 = c 2 n &#8722; 1 &#8658; 2 n c + n &#8722; 1 = c 2 n &#8722; 1 &#8658; c 2 n &#8722; 2 n &#8722; 1 = 2 n 2 &#8722; 2 n &#8658; c = 2 n 2 &#8722; 2 n 2 n &#8722; 2 n &#8722; 1 &#8805; 1 &#160; . . . . . . . . . . . . . i &#8658; 2 n 2 &#8722; 2 n &#8805; 2 n &#8722; 2 n &#8722; 1 &#8658; 2 n 2 + 1 &#8805;

Question 9

Let A P &#160; a ; d denote the set of all the terms of an infinite arithmetic progression with first term a and common difference d &#62; 0 . If A P 1 ; 3 &#8745; A P 2 ; 5 &#8745; A P 3 ; 7 = A P a

Accepted Answer

Let T 1 , &#160; m : m t h term of 1 s t series A 1 ; 3 T 2 , &#160; n : n t h term of 2 n d series A 2 ; 5 T 3 , &#160; r : r t h term of 3 r d series A ( 3 ; 7 ) For common terms T ( 1 , &#160; m ) = T ( 2 , &#160; n ) = T ( 3 , &#160; r ) 1 + 3 m - 1 = 2 + 5 n - 1 = 3 + 7 r - 1 3 m - 2 = 5 n - 3 = 7 r - 4 &#8658; m = 5 n - 1 3 &#8201;&#8201;&#8201; a n d &#8201;&#8201;&#8201; r = 5 n + 1 7 For &#8216; m &#8217; to be a natural number n = 2,5 , 11 , &#8230; And for &#8216; r &#8217; to be a na

Sequences and Series — Previous Year Questions & Solutions