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QUESTION: 1

The sum of integers from 1 to 100 that are divisible by 2 or 5 is

Solution:

Set of numbers divisible by 2 are 2,4,6,...,100

Set of numbers divisible by 5 are 5,10,15,...,100

Set of numbers divisible by 10 are 10,20,30,...,100

Now, sum of numbers divisible by 2 are given by

Sum of numbers divisible by 5 are given by

Sum of numbers divisible by 10 are given by

Hence required sum = S_{2} +S_{5} −S_{10 }= 2550 + 1050 − 550 = 3050

QUESTION: 2

Consider an A.P. with first term `a' and the common difference `d'. Let S_{k} denote the sum of its first K terms. If is independent of x, then

Solution:

Given, first term =a, common difference =d

By using Sn = n/2 [2a+(n−1)d] we have,

∴ Skx / Sx = kx/2(2a+(kx−1)d) / x/2(2a+(x+1)d)

⇒Skx / Sx = k(2a+(kx−1)d) / (2a+(x+1)d)

⇒Skx / Sx = k(2a−d+kxd) / 2a+xd−d

Now this will be independent of x only when 2a=d or a = d/2

QUESTION: 3

The next term of the sequence, 2, 6, 12, 20, …..is

Solution:

nth term of given sequence is n(n+1)

so the 5th term is 5(5+1)=30

QUESTION: 4

If A, G & H are respectively te A.M., G.M. & H.M. of three positive numbers a, b, & c then the equation whose roots are a, b & c is given by

Solution:

QUESTION: 5

If a^{x} = b^{y} = c^{z} = d^{t}_{ }and a, b, c, d are in G.P., then x, y, z, t are in

Solution:

QUESTION: 6

If x_{i} > 0, i = 1, 2,....,50 ans x_{1} + x_{2} + ....+x_{50} = 50, then the minimum value of + + ......+equal to

Solution:

QUESTION: 7

If a, a_{1}, a_{2}, a_{3}, ....... , a_{2n}, b are in A.P. and a, g_{1}, g_{2}, g_{3},.....g_{2n}, b are in G.P. and h is the harmonic mean of a and b, then + +........ .....+ , is equal to

Solution:

QUESTION: 8

One side of an equilateral triangle is 24 cm. The mid-points of its sides are joined to form another triangle whose mid-points are in turn joined to form still another triangle. This process continues indefinitely. Then the sum of the perimeters of all the trianlges is

Solution:

QUESTION: 9

If a_{1}, a_{2}....a_{n} are in A.P. with common difference d > 0, then the sum of the series (sin d) [cosec a_{1} cosec a_{2} + cosec a_{2} cosec a_{3} +..... ......+ cosec an – 1 cosec a_{n}]

Solution:

A₁ , A₂ , A₃ , .......An are in AP and common difference is d

so, d = A₂ - A₁ = A₃ - A₂ = A₄ - A₃ =...... =An - A_{n-1} -------(1)

Now, sind[cosecA₁.cosecA₂ + cosecA₂.cosecA₃ + cosecA₃.cosecA₄ ...........+ cosecA_{n-1}.cosecAn ]

⇒ sind[1/sinA₁.sinA₂ + 1/sinA₂.sinA₃ + 1/sinA₃.sinA₄ + .........+ 1/sinAn₋₁.sinAn]

⇒ sin d/sinA₁.sinA₂ + sind/sinA₂.sinA₃ + sind/sinA₃.sinA₄ + .......+sind/sinAn₋₁.sinAn

⇒ sin(A₂ - A₁)/sinA₁.sinA₂ + sin(A₃ - A₂)/sinA₂.sinA₃ + sin(A₄ - A₃)/sinA₃.sinA₄ + .......+ sin(A_{n-1} - An)/sinAn-1.sinAn [ from equation (1)

⇒ (sinA₁.cosA₂ - cosA₁.sinA₂)/sinA₁.sinA₂ + (sinA₂.cosA₃ - cosA₂.sinA₃)/sinA₂.sinA₃ +.........+ (sinA_{n-1}.cosAn - cosA_{n-1}.sinA_{n})/sinA_{n-1}.sinA_{n}

⇒cotA₂ - cotA₁ + cotA₃ - cotA₂ + cotA₄ - cotA₃ + .........cotA_{n} - cotA_{n}-1

⇒cotA_{n} - cotA₁

QUESTION: 10

An infinite GP has first term x and sum 5, then x belongs to

Solution:

QUESTION: 11

If S_{1}, S_{2}, S_{3} are the sums of first n natural numbers, their squares, their cubes respectively, thenis equal to

Solution:

QUESTION: 12

If a_{1}, a_{2}, ...., a_{n} are in HP, then the expression a_{1}a_{2} + a_{2}a_{3} + ....+ an – 1 a_{n} is equal to

Solution:

QUESTION: 13

If x^{2} + 9y^{2} + 25z^{2} = xyz, then x, y and z are in

Solution:

QUESTION: 14

The sum of all possible products of first n natural numbers taken two by two is

Solution:

QUESTION: 15

If G_{1} and G_{2} and two geometric means and A is the arithmetic means inserted between two positive numbers then the value of is

Solution:

QUESTION: 16

{a_{n}} and {b_{n}} are two sequences given bya_{n} = and b_{n} = for all n ∈ N. The value of a_{1} a_{2} a_{3}...... a_{n} is equal to

Solution:

QUESTION: 17

If 1^{2} + 2^{2} + 3^{2} + ..... + 2003^{2} = (2003) (4007) (334) and (1) (2003) + (2) (2002) + (3) (2001) +.... ....+ (2003) (1) = (2003) (334) (x)., then x equals

Solution:

for 2003 terms the summation is = (2003)(2004)(4007) / 6 = (2003)(4007)(334)

Now,

(1)(2003)+(2)(2002)+(3)(2001)+....(2003)(1) =

= [ 2004(2003)(2003 + 1) / 2 ] - [ (2003)(4007)(334) ]

= [ (1002)(2003)(2004) ] - [ (2003)(4007)(334) ]

= 3 [(334)(2003)(2004) ] - [ (2003)(4007)(334)]

= (2003)(334)( 3*2004 - 4007)

= (2003)(334)(2005)

So the answer is 2005.

QUESTION: 18

Sum of first fifteen terms of series 5+10+20+… is

Solution:

a= 5 r=2

ar^{n-1} = 5(2)^{n-1}

Sn = a(r^{n -1})/r-1

= 5(2^{15-1})/2-1

= 5(32768-1)1

= 98301

QUESTION: 19

The common difference d of the A.P. in which T_{7} = 9 and T_{1}T_{2}T_{7} is least is

Solution:

QUESTION: 20

The H.M. between two numbers is , their A.M. is A and G.M. is G. If 2A + G^{2} = 26, then the numbers are

Solution:

QUESTION: 21

If 1, 2, 3.... are first terms; 1, 3, 5..... are common differences and S_{1}, S_{2}, S_{3}.... are sums of n terms of given p AP's; then S_{1} + S_{2} + S_{3} + .... + S_{p} is equal to

Solution:

*Multiple options can be correct

QUESTION: 22

For the A.P. given by a_{1}, a_{2},...............a_{n},..........., the equations satisfied are

Solution:

*Multiple options can be correct

QUESTION: 23

If sum of the infinite G.P., p,1,1/p^{2} ,1/p^{3},...... is 9 /2, the value of p is

Solution:

*Multiple options can be correct

QUESTION: 24

If positive numbers a, b, c are in A.P. and a^{2},^{ }b^{2}, c^{2} are in H.P., then

Solution:

*Multiple options can be correct

QUESTION: 25

If the arithmetic mean of two positive numbers a & b (a > b) is twice their geometric mean, then a : b is

Solution:

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