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QUESTION: 1

Let a, b, c, be in A.P. ,with a common difference d, then e^{1/e} , e^{b}^{/ae}, e^{1/a} are in

Solution:

QUESTION: 2

Sum to 20 terms of the series (1.3)^{2} + (2.5)^{2} + (3.7)^{2} +… is:

Solution:

Given that (1.3)^{2}, (2.5)^{2} , (3.7)^{2} .....................20 terms

Here 1,2,3,.................are in A.P.

a = 1, d = 1

t_{n} = a + (n-1) d = 1 + (n -1)1 = n

3,5,7 ............. are in A.P

a = 3, d = 2

tn = a + (n-1) d = 3 + (n -1)^{2} = 2n + 1

∴ nth term = n(2n + 1)^{2} = 4n^{3} + 4n^{2} + n

Sum of n terms Sn = ∑ ( 4n^{3} + 4n^{2} + n ) = 4 { n^{2}(n+1)^{2}}/4 + 4 {n (n+1)(2n+1)} / 6 + n(n+1)/2

= { n^{2}(n+1)^{2}} + 2 {n (n+1)(2n+1)} / 3 + n(n+1)/2

Sum of 20 terms S_{20} = 400 × 441 + 40 × 41 × 7 + 10 ×21 = 176400 + 11480 + 210

= 188090.

QUESTION: 3

0.2 + 0.22 + 0.222 + ……. to n terms =

Solution:

0.2 + 0.22 + 0.222 + 0.2222 + ….

2(0.1 + 0.11 + 0.111 + 0.1111 + ….)

2/9 (0.9 + 0.99 + 0.999 + 0.9999 + ….)

2/9 (( 1 - 0.1 ) + ( 1 - 0.01 ) + ( 1 - 0.001 ) + ( 1 - 0.0001 ) + …. )

2/9 (1 - 0.1 + 1 - 0.01 + 1 - 0.001 + 1 - 0.0001 + ….)

2/9 ((1 + 1 + 1 + 1 + ….) - ( 0.1 + 0.01 + 0.001 + 0.0001 + ….))

2/9 (n - ( 0.1 + 0.01 + 0.001 + 0.0001 + ….))

2/9 (n - (( 0.1 ( 1 - 0.1^{n})/( 1 - 0.1))))

2/9 (n - (( 0.1 / 0.9)( 1 - 0.1^{n})))

2/9 (n - (1/9)(1 - 0.1^{n}))

QUESTION: 4

The sum of the series 3^{2} + 6^{2} + 9^{2} + … + (3n)^{2} is:

Solution:

3^{2} + 6^{2 }+....

∑(3n)^{2}

9∑n^{2}

9n(n+1)(2n+1)/6

= 3n(n+1)(2n+1)/2

QUESTION: 5

In how many different ways the letters of the word HEXAGON can be permuted?

Solution:

The given word consists of 7 different letters out of which 3 are vowels and 4 are consonants.

7 different letters can be arranged in

7! ways = 5040 ways.

QUESTION: 6

The sum of the squares of the first 15 natural numbers is

Solution:

S_{n} = 1/6n(n+1)(2n+1)

Here n = 15 so,

S_{15} = 1/6 × 15(15+1)(2×15+1)

S_{15} = 1/6 × 15 × 16(30 + 1)

S_{15} = 1/6 × 15 × 16 × 31

S_{15} = 1240

QUESTION: 7

Sum of squares of the first n natural numbers is given by:

Solution:

The formula to find sum of first n terms of a sequence of squares of natural numbers = n(n+1)(2n+1)/6

QUESTION: 8

The sum of the series 1^{3} – 2^{3} + 3^{3} – 4^{3} + ……. + 9^{3} is:

Solution:

1^{3} + 3^{3} + 5^{3} + 7^{3} + 9^{3} − (2^{3} + 4^{3} + 6^{3} + 8^{3})

= Sum of cubes of first 5 odd number -Sum of cubes of first 4 even number

= n^{2}(2n^{2}–1) − 2n^{2}(n+1)^{2}

= 1225 – 800

= 425

QUESTION: 9

Sum to n terms of the series 2 + 5 + 14 + 41 +… is:

Solution:

S_{n} = 2 + [2+3^{1}] + [2+3^{1}+3^{2}] + ⋯ + [2+3^{1}+3^{2}+⋯+3^{n-1}]

= 2n+[3^{1}]+[3^{1}+3^{2}]+⋯+[3^{1}+3^{2}+⋯+3^{n-1}]

= 2n + t_{3,1} + t_{3,2} +⋯+t_{3,n-1}

= 2n+T_{3,n-1}

= 2n+3/4(3^{n}−2n−1)

= n/2 + 3/4(3^{n} − 1)

QUESTION: 10

If 1, a and P are in A. P. and 1, g and P are in G. P., then

Solution:

2a = 1 + p and g^{2} = P

⇒ g^{2} = 2a − 1

⇒ 1 − 2a + g^{2} = 0

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