Question

There are m multiples of 6 in range [0, 100] and n multiples of 6 in [-6, 35]. Find out the value of X, if X = m - n. Where X, m, and n are positive integer.

Answer 1

Answer:

10

Step-by-step explanation:

Using the indentities of AP:

xth term = a + (x - 1)d, a is the first term and d is the common difference.

For m:

First no. divisible by 6 is 6 and last term is 96, in [0, 100].

First term = 0, mth term = 96, d = 6.

=> mth term = a + (m - 1)d

=> 96 = 0 + (m - 1)(6)

=> 17 = m

For n:

First no. divisible by 6 is -6 and last term is 30, in [-6, 35].

First term = -6, nth term = 30, d = 6.

=> nth term = a + (n - 1)d

=> 30 = -6 + (n - 1)(6)

=> 7 = n

Therefore,

X = m - n => X = 17 - 7

X = 10

If you're not familiar to AP:

If you start with 6, you would stop at 96, if you keep on adding 6 till 100(from 0).

You get: terms = 17

m = 17

Similarly, if you start from -6 and keep on adding 6 till 35, you get n = 7.

Hence, X = 17 - 7 = 10

Answer 2

Answer:

Given :-

• There are m multiple of 6 in range of [0, 100] and n multiple of 6 in [- 6, 35].
• m and n are positive integers.

To Find :-

• What is the value of X, if X = m - n.

Solution :-

${\small{\bold{\purple{\underline{\bigstar\: In\: case\: of\: m\: multiple\: :-}}}}}$

Given :

• First term (a) = 0
• mth term of an AP ($\sf a_m$) = 96
• Common difference (d) = 6

As we know that,

$\clubsuit$ nth term of an AP Formula :

$\mapsto \sf\boxed{\bold{\pink{a_n =\: a + (n - 1)d}}}$

where,

• $\sf a_n$ = nth term of an AP
• a = First term of an AP
• n = Number of terms
• d = Common difference

Similarly,

$\clubsuit$ mth term of an AP Formula :

$\mapsto \sf\boxed{\bold{\pink{a_m =\: a + (m - 1)d}}}$

where,

• $\sf a_m$ = mth term of an AP
• a = First term of an AP
• m = Number of terms
• d = Common difference

According to the question by using the formula we get,

$\implies \sf 96 =\: 0 + (m - 1)6$

$\implies \sf 96 =\: 0 + 6m - 6$

$\implies \sf 96 =\: 6m - 6$

$\implies \sf 96 + 6 =\: 6m$

$\implies \sf 102 =\: 6m$

$\implies \sf \dfrac{\cancel{102}}{\cancel{6}} =\: m$

$\implies \sf \dfrac{17}{1} =\: m$

$\implies \sf 17 =\: m$

$\implies \sf\bold{\green{m =\: 17}}$

${\small{\bold{\purple{\underline{\bigstar\: In\: case\: of\: n\: multiple\: :-}}}}}$

Given :

• First term (a) = - 6
• nth term of an AP ($\sf a_n$) = 30
• Common difference (d) = 6

So, for nth term we know that :

$\mapsto \sf\boxed{\bold{\pink{a_n =\: a + (n - 1)d}}}$

where,

• $\sf a_n$ = nth term of an AP
• a = First term of an AP
• n = Number of terms
• d = Common difference

According to the question by using the formula we get,

$\implies \sf 30 =\: - 6 + (n - 1)6$

$\implies \sf 30 =\: - 6 + 6n - 6$

$\implies \sf 30 =\: - 6 - 6 + 6n$

$\implies \sf 30 =\: - 12 + 6n$

$\implies \sf 30 + 12 =\: 6n$

$\implies \sf 42 =\: 6n$

$\implies \sf \dfrac{\cancel{42}}{\cancel{6}} =\: n$

$\implies \sf \dfrac{7}{1} =\: n$

$\implies \sf 7 =\: n$

$\implies \sf \bold{\green{n =\: 7}}$

Hence, the value of X will be :

We have :

$\bullet\: \: \sf\bold{m =\: 17}$

$\bullet\: \: \sf\bold{n =\: 7}$

Then, the value of X is :

$\longrightarrow \sf X =\: m - n$

$\longrightarrow \sf X =\: 17 - 7$

$\longrightarrow \sf\bold{\red{X =\: 10}}$

$\therefore$ The value of X is 10.