Q: If _ k=1 ^n k(k+1)(k-1)=p n^4+q n^3+t n^2+s n where p, q, t and s are constants, then the value of s is equal to

aligned Let S & = _ k=1 ^n k(k+1)(k-1) \\ & = _ k=1 ^n k^3-k aligned We know that _ r=1 ^P r= P(P+1) 2 and _ r=1 ^P r^3= [ P(P+1) 2 ]^2 aligned S & = [ n(n+1) 2 ]^2- [ n(n+1) 2 ] \\ & = n(n+1) 2 [ n(n+1) 2 -1 ] \\ & = ( n^2+n 2 ) [ n^2+n-2 2 ] \\ & = n^4 4 + n^3 2 - n^2 4 - n 2 ...(i) aligned Now, by comparing p n^4+q n^3+t n^2+s n with Eq. (i), S=- 1 2

Q: If ^ -1 [ 1 1+1.2 ]+ ^ -1 [ 1 1+2.3 ]+ + ^ -1 [ 1 1+n(n+1) ]= ^ -1 [x], then x is equal to

aligned & Given ^ -1 ( 1 1+1.2 )+ ^ -1 ( 1 1+2.3 )+ \\ & + ^ -1 ( 1 1+n(n+1) )= ^ -1 (x) aligned aligned ^ -1 ( 2-1 1+1.2 ) & + ^ -1 ( 3-2 1+2.3 )+ \\ & + ^ -1 ( n+1-n 1+n(n+1) )= ^ -1 (x) aligned aligned & ^ -1 (2)- ^ -1 (1)+ ^ -1 (3)- ^ -1 (2)+ \\ & + ^ -1 ( n +1)- ^ -1 ( n )= ^ -1 (x) aligned ^ -1 ( n +1)- ^ -1 (1)= ^ -1 (x) \ ^ -1 (A)- ^ -1 (B)= ^ -1 ( A-B 1+A B ) \ ^ -1 ( n n+2 )= ^ -1 (x) x= n n+2

Question 1

If a 0, b 0, c 0 and a, b, c are distinct, then (a+b)(b+c)(c+a) is greater than

Accepted Answer

We know that AM GM ...(i) Now, applying Eq. (i) in a and b a+b 2 a b ....(ii) applying Eq. (i) in b and c b+c 2 b c ....(iii) and applying Eq. (i) in c and a c+a 2 c a ....(iv) By E'qs. (ii), (iii) and (iv) array cc & (a+b)(b+c)(c+a) 8 a b c \ & (a+b)(b+c)(c+a) 8 a b c array

Question 2

If _ k=1 ^n k(k+1)(k-1)=p n^4+q n^3+t n^2+s n where p, q, t and s are constants, then the value of s is equal to

Accepted Answer

aligned Let S & = _ k=1 ^n k(k+1)(k-1) \ & = _ k=1 ^n k^3-k aligned We know that _ r=1 ^P r= P(P+1) 2 and _ r=1 ^P r^3= [ P(P+1) 2 ]^2 aligned S & = [ n(n+1) 2 ]^2- [ n(n+1) 2 ] \ & = n(n+1) 2 [ n(n+1) 2 -1 ] \ & = ( n^2+n 2 ) [ n^2+n-2 2 ] \ & = n^4 4 + n^3 2 - n^2 4 - n 2 ...(i) aligned Now, by comparing p n^4+q n^3+t n^2+s n with Eq. (i), S=- 1 2

Question 3

There are four numbers of which the first three are in GP and the last three are in AP, whose common difference is 6 . If the first and the last numbers are equal, then two other numbers are

Accepted Answer

Let 3 numbers in AP are a,(a+6),(a+12) Also, first and last number out of 4 numbers are equal 4 numbers are (a+12), a,(a+6),(a+12) Also, given, first 3 numbers are in GP array ll & a^2=(a+12)(a+6) \ & a^2=a^2+18 a+72 \ & a=- 72 18 =-4 array 4 numbers are 8,-4,2,8

Question 4

If A=1+r^a+r^ 2 a +r^ 3 a + and B =1+ r ^ b + r ^ 2 ~b + r ^ 3 ~b + , then a b is equal to

Accepted Answer

A= 1 1-r^a 1-r^a= 1 A r^a=1- 1 A = A-1 A B= 1 1-r^b 1-r^b= 1 B r^b=1- 1 B = B-1 B a r= ( A-1 A ) and b r= ( B-1 B ) a b = ( A -1 ~A ) ( B -1 ~B ) = _ B -1 ~B ( ~A -1 ~A )

Question 5

The sum of the infinite series 1+ 5 6 + 12 6^2 + 22 6^3 + 35 6^4 + 51 6^5 + 70 6^6 + .. is equal to:

Accepted Answer

Let S=1+ 5 6 + 12 6^2 + 22 6^3 + 35 6^4 + .....(i) S 6 = 1 6 + 5 6^2 + 12 6^3 + 22 6^4 + ....(ii) On subtracting (i) from (ii) 5 6 S=1+ 4 6 + 7 6^2 + 10 6^3 + 13 6^4 + 5 36 S= 1 6 + 4 6^2 + 7 6^3 + 10 6^4 + 13 6^5 + on subtraction 25 36 S=1+ 3 6 + 3 6^2 + 3 6^3 + = 8 5 S= 8 5 36 25 = 288 125

Question 6

If ^ -1 [ 1 1+1.2 ]+ ^ -1 [ 1 1+2.3 ]+ + ^ -1 [ 1 1+n(n+1) ]= ^ -1 [x], then x is equal to

Accepted Answer

aligned & Given ^ -1 ( 1 1+1.2 )+ ^ -1 ( 1 1+2.3 )+ \ & + ^ -1 ( 1 1+n(n+1) )= ^ -1 (x) aligned aligned ^ -1 ( 2-1 1+1.2 ) & + ^ -1 ( 3-2 1+2.3 )+ \ & + ^ -1 ( n+1-n 1+n(n+1) )= ^ -1 (x) aligned aligned & ^ -1 (2)- ^ -1 (1)+ ^ -1 (3)- ^ -1 (2)+ \ & + ^ -1 ( n +1)- ^ -1 ( n )= ^ -1 (x) aligned ^ -1 ( n +1)- ^ -1 (1)= ^ -1 (x) \ ^ -1 (A)- ^ -1 (B)= ^ -1 ( A-B 1+A B ) \ ^ -1 ( n n+2 )= ^ -1 (x) x= n n+2

Question 7

If arithmetic mean of two distinct positive real number a and b(a b) be twice their geometric mean, then a: b=

Accepted Answer

By the given condition, a+b 2 =2 a b a+b=4 a b Now, (a-b)^2=(a+b)^2-4 a b=16 a b-4 a b= 12 a b a-b= 12 a b =2 3 a b . (Taking + ve sign only as a b) a+b a-b = 4 a b 2 3 a b = 2 3 By componendo and dividendo, 2 a 2 b = 2+ 3 2- 3 or a b = 2+ 3 2- 3

Question 8

If y= ^ -1 1 x^2+x+1 + ^ -1 1 x^2+3 x+3 + ^ -1 1 x^2+5 x+7 + .to n terms, then d y d x =

Accepted Answer

Given, y= ^ -1 1 x^2+x+1 + ^ -1 1 x^2+3 x+3 + ^ -1 1 x^2+5 x+7 + to n terms = ^ -1 \ 1 1+x(x+1) \ + ^ -1 \ 1 1+(x+1)(x+2) \ + ^ -1 \ 1 1+(x+2)(x+3) \ + + ^ -1 \ 1 1+(x+(n-1))(x+n) \ = ^ -1 \ (x+1)-x 1+(x+1) x \ + ^ -1 \ (x+2)-(x+1) 1+(x+2)(x+1) \ + ^ -1 \ (x+3)-(x+2) 1+(x+3)(x+2) \ + .+ ^ -1 ( (x+n)-(x+n-1) 1+(x+n)(x+n-1) ) y= \ ^ -1 (x+1)- ^ -1 (x) \ + \ ^ -1 (x+2)- ^ -1 (x+1) \ + \ ^ -1 (x+3)- ^ -1 (x+2) \ + . .+ \ ^ -1 (x+n)- ^ -1 [x+(n-1)] \ So, y= ^ -1 (x+n)- ^ -1 (x) On differentiating bot

Question 9

The sum to infinite term of the series 1+ 2 3 + 6 3^2 + 10 3^3 + 14 3^4 + is

Accepted Answer

We have Multiplying both sides by 1 3 , we get Subtracting eqn. (ii) from eqn. (i), we get aligned & 2 3 S=1+ 1 3 + 4 3^2 + 4 3^3 + 4 3^4 + \ & 2 3 S= 4 3 + 4 3^2 + 4 3^3 + 4 3^4 + \ & 2 3 S= 4 3 1- 1 3 = 4 3 3 2 S=3 aligned

Question 10

If the sum of an infinite GP a, a r, a r^2, a r^3, is 15 and the sum of the squares of its each term is 150 , then the sum of a r^2, a r^4, a r^6, is :

Accepted Answer

Sum of infinite terms of series a +a r+a r^2+ =15 a 1-r =15 ...(i) Sequence formed by square of terms: a^2, a^2 r^2, a^2 r^4, a^2 r^6 a^2+a^2 r^2+ =150 Sum = a^2 1-r^2 =150 a 1-r a 1+r =15 15 a 1+r =150 ( a 1- r =15 ) a 1+r =10 ...(ii) On dividing equation (i) by (ii) 1+ r 1- r = 15 10 or r= 1 5 a=12 Now series : a r^2+a r^4+a r^6+ Sum = a^2 1-r^2 = 12 ( 1 25 ) 1- 1 25 = 1 2

Question 11

Let a_1, a_2, a_ 40 be in AP and h_1, h_2, . h_ 10 be in HP. If a_1=h_1=2 and a_ 10 =h_ 10 =3, then a_4 h_7 is

Accepted Answer

Let d be the common difference of the AP. Then, aligned & a_ 10 =3 a_1+9 d=3 \ & 2+9 d=3 d= 1 9 \ & a_4=a_1+3 d=2+ 1 3 = 7 3 aligned Let D be the common difference of 1 h_1 , 1 h_2 , .- 1 h_ 10 . Then, h_ 10 =3 array ll & 1 h_ 10 = 1 3 1 2 +9 D= 1 3 \ & 9 D=- 1 6 D=- 1 54 \ & 1 h_7 = 1 h_1 +6 D= 1 2 - 1 9 = 7 18 \ & h_7= 18 7 \ & a_4 h_7= 7 3 18 7 =6 array

Question 12

Let a_1, a_2, a_3 be a harmonic progression with a_1=5 and a_ 20 =25. The least positive integer n for which a_n < 0, is

Accepted Answer

It is given that a_1, a_2, a_3 . are in HP. Therefore, 1 a_1 , 1 a_2 , 1 a_3 , .. are in AP. Let d be the common difference of the AP. 1 a_n = 1 a_1 +(n-1) d and 1 a_ 20 = 1 a_1 +19 d 1 a_n = 1 5 +(n-1) d and 1 25 = 1 5 +19 d 1 a_n = 1 5 +(n-1) d and d= -4 19 25 1 a_n = 1 5 - 4(n-1) 19 25 1 a_n = 95-4 n+4 19 25 a_n= 99-4 n 19 25 Now, a_n 99 n>24 3 4 n 25

Question 13

In a sequence of 21 terms, the first 11 terms are in AP with common difference 2 and the last 11 terms are in GP with common ratio 2. If the middle term of AP be equal to the middle term of the GP, th

Accepted Answer

Since, the first 11 terms are in AP, d=2 a_ 11 =a+10 d=a+20 The middle term of AP is T_6=a+5 d=a+10 For the next 11 terms in GP r=2 The middle term of GP is b(2)^5 where, b is the first term of a GP which is the last term of AP b(2)^5=(a+20) 32 According to the given condition, aligned a+10 & =(a+20) 32 \ 31 a & =10-640 \ a & =- 630 31 aligned Middle term of entire sequence is 11th term aligned T_ 11 & = -630 31 +10 d \ & = -630 31 +10 2= -10 31 aligned

Question 14

If , , [0, ] and if , , are in AP, then - - is equal to

Accepted Answer

- - = 2 + 2 - 2 2 + 2 - 2 = + 2 [ array l A- B=2 ( A+B 2 ) ( A-B 2 ) \ and A- B=2 ( A+B 2 ) ( B-A 2 ) array ] But , , are in AP + 2 = So, required value is .

Question 15

If the sum of a certain number of terms of the A.P. 25,22,19, is 116 then the last term is

Accepted Answer

a=25, d=22-25=-3 . Let n be the no. of terms array l Sum =116 ; \ Sum = n 2 [2 a+(n-1) d] \ 116= n 2 [50+(n-1)(-3)] \ or 232=n[50-3 n+3]=n[53-3 n]=-3 n^ 2 +53 n \ 3 n^ 2 -53+232=0 \ (n-8)(3 n-29)=0 \ n=8 or n= 29 3 , \ n 29 3 \ n=8 \ Now, T _ 8 =a+(8-1) d=25+7 (-3)=25-21=4 \ Last term =4 array

Question 16

The sum of all odd numbers between 1 and 1000 which are divisible by 3 is

Accepted Answer

Sum of odd numbers between 1 and 1000 , which is divisible by 3=3+9+15+21+27+ +999= S (let) Let n be the number of terms in series and a is first term. l=a+(n-1) d, where l is last term and d is is common difference. array l 999=3+(n-1) 6 \ n-1= 999-3 6 = 996 6 \ n-1=166 \ n=167 \ S = n 2 [2 a+(n-1) d] \ = 167 2 [2 3+(167-1) 6] \ = 167 2 [1002]=167 501=83667 array

Question 17

Consider an infinite geometric series with first term a and common ratio r. If its sum is 4 and the second term is 3 4 , then:

Accepted Answer

Since, sum =4 and second term = 3 4 a 1-r =4, and a r= 3 4 a 1- 3 4 a =4 (a-1)(a-3)=0 a=1 or a=3 array l a 1- 3 4 a =4 \ (a-1)(a-3)=0 \ a=1 or a=3 array

Question 18

If the harmonic mean between a and b be H, then the value of 1 H -a + 1 H -b is

Accepted Answer

Putting H = 2 a b a+b , we have array l 1 H -a + 1 H -b \ = 1 ( 2 a b a+b -a ) + 1 ( 2 a b a+b -b ) = a+b a b-a^ 2 + a+b a b-b^ 2 array = ( a+b b-a ) ( 1 a - 1 b )= ( a+b b-a ) ( b-a a b )= a+b a b = 1 a + 1 b

Question 19

If a, b, c are positive numbers, then least value of (a+b+c) ( 1 a + 1 b + 1 c ) is

Accepted Answer

AM HM a+b+c 3 3 1 a + 1 b + 1 c (a+b+c) ( 1 a + 1 b + 1 c ) 9

Question 20

The sum of the infinite series 2^ 2 2 ! + 2^ 4 4 ! + 2^ 6 6 ! + is equal to

Accepted Answer

We know that e^ x +e^ -x 2 =1+ x^ 2 2 ! + x^ 4 4 ! + x^ 6 6 ! + keeping x =2, we get Expression = 1 2 [ e^ 2 +e^ -2 2 ]-1= (e^ 2 -1 )^ 2 2 e^ 2

Sequences and Series — Previous Year Questions & Solutions