Question

✔✔$OLVE TH€ @BOVE QUE$TION.....

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Answer 1

Answer:

i wrote the answer with detailed explanation in the above image

Answer 2

ANSWER:5

Let a= n³ -n

=) a= n(n² -1)

=) a=n(n-1) × (n-1) [Using (a² - b²)]

=) a= (n-1)× n× (n+1)

We know that,

1.If a number is completely divisible by 2 & 3, then it is also divisible by 6.

2.If the sum of digits of any number is divisible by 3, then it's also divisible by 3.

3.If one of the factors of any number is an even number, then it's also divisible by 2.

So,

a=(n-1)×n × (n+1)

Sum of the digits:

=) n-1+n+n+1

=) 3n which is a multiple of 3, where n is any positive integer.

&

(n-1)× n ×(n+1) will always be even, as one out of (n-1) or n or (n+1) be even integers.

Hence,

By condition 1 the number is n³ - n is always divisible by 6, where n is any positive Integer.

ANSWER:6

Given:

A shopkeeper buys a number of books for Rs.80. If he had bought 4 more books for the same amount, each book would have cost Rs.1 less.

To find:

How many books did he buy?

Solution:

Let the number of books bought be x.

So,

•Cost of x books= Rs.80

=) cost of 1 book= Rs.80/x

If the number of books bought is x+4.

=) Cost of one book= Rs.80/x+4

According to the question:

Therefore,

$= > \frac{80}{x} - \frac{80}{x + 4} = 1 \\ \\ = > 80( \frac{1}{x} - \frac{1}{x + 4} ) = 1 \\ \\ = > 80( \frac{x + 4 - x}{x(x + 4)} ) = 1 \\ \\ = > 80( \frac{4}{ {x}^{2} + 4x } ) = 1 \\ \\ = > \frac{320}{ {x}^{2} + 4x} = 1 \\ \\ = > {x}^{2} + 4x = 320 \\ \\ = > {x}^{2} + 4x - 320 = 0 \\ \\ = > {x}^{2} + 20x - 16x - 320 = 0 \\ \\ = > x(x + 20) - 16(x + 20) = 0 \\ \\ = > (x + 20)(x - 16) = 0 \\ \\ = > x + 20 = 0 \: \: \: or \: \: \: x - 16 = 0 \\ \\ = > x = - 20 \: \: \: or \: \: \: x = 16$

So,

x can't be negative.

Acceptable value is x= 16.

Hence,

He buy number of books is 16.

ANSWER:7

Given:

The sum of first six term of an A.P is 42.

The ratio of 10th term to its 30th term is 1:3.

To find:

The first term of the A.P. & 13th term.

Solution:

Let a is the first term & d is the common difference in an A.P.

We know that, formula of A.P.

⚫nth term= an= a + (n-1)d

&

Sum of first n terms of an A.P.

$= > {}^{S} n = \frac{n}{2} [2a + (n - 1)d]$

Now,

Sum of First 6th terms, S6= 42.

$= > \frac{6}{2} [2a + (6 - 1)d] = 42 \\ \\ = > 3(2a + 5d) = 42 \\ \\ = > 2a + 5d = 14.................(1)$

&

Given ratio of 10th term & 30th term is 1:3.

$= > \frac{ {}^{a}10 }{ {}^{a} 30} = \frac{1}{3} \\ \\ = > \frac{a + 9d}{a + 29d} = \frac{1}{3} \\ [Cross \: multiplication] \\ \\ = > 3a + 27d = a + 29d \\ \\ = > 3a - a = 29d - 27d \\ \\ = > 2a = 2d \\ \\ = > a = d$

So,

Putting the value of d=a in equation (1), we get;

=) 2a + 5(a)= 14

=) 2a + 5a= 14

=) 7a = 14

=) a= 14/7

=) a= 2

Therefore,

•First term of A.P. is 2.

&

13th term of an A.P.

=) a + (n-1)d.

=) 2 + (13-1)2

=) 2 + 12×2

=) 2 + 24

=) 26

Hence,

⚫The first term of an A.P. is 2.

⚫13th term of an A.P. is 26.

Hope it helps ☺️