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18.

(a) If the [tex]p^{\text{th}}[/tex] term of an A.S. is [tex]q[/tex] and the [tex]q^{\text{th}}[/tex] term is [tex]p[/tex], show that the [tex]m^{\text{th}}[/tex] term is [tex]p+q-m[/tex].

(b) If [tex]m[/tex] times the [tex]m^{\text{th}}[/tex] term of an A.P. is equal to [tex]n[/tex] times its [tex]n^{\text{th}}[/tex] term, show that...

Sagot :

GinnyAnsweridn
Sure, let's go through the steps to solve the problem clearly.

### Part (a)

Given:
1. The [tex]\( p \)[/tex]-th term of an arithmetic sequence (A.S) is [tex]\( q \)[/tex].
2. The [tex]\( q \)[/tex]-th term is [tex]\( p \)[/tex].

We need to show that the [tex]\( m \)[/tex]-th term is [tex]\( p + q - m \)[/tex].

Let's denote the first term of the arithmetic sequence by [tex]\( a \)[/tex] and the common difference by [tex]\( d \)[/tex].

The general formula for the [tex]\( n \)[/tex]-th term of an A.S. is:
[tex]\[ T_n = a + (n-1)d \][/tex]

Given:
1. [tex]\( T_p = q \)[/tex]
2. [tex]\( T_q = p \)[/tex]

Using the formula for the [tex]\( p \)[/tex]-th term:
[tex]\[ a + (p-1)d = q \tag{1} \][/tex]

Using the formula for the [tex]\( q \)[/tex]-th term:
[tex]\[ a + (q-1)d = p \tag{2} \][/tex]

Now, subtract equation (1) from (2):
[tex]\[ (a + (q-1)d) - (a + (p-1)d) = p - q \][/tex]
[tex]\[ (q-1)d - (p-1)d = p - q \][/tex]
[tex]\[ (q-p)d = p - q \][/tex]
[tex]\[ d = \frac{p-q}{q-p} \][/tex]
[tex]\[ d = -1 \][/tex]

So, we find that [tex]\( d = -1 \)[/tex].

Now substitute [tex]\( d = -1 \)[/tex] back into equation (1):
[tex]\[ a + (p-1)(-1) = q \][/tex]
[tex]\[ a - (p-1) = q \][/tex]
[tex]\[ a = q + p - 1 \][/tex]

We need to find the [tex]\( m \)[/tex]-th term:
[tex]\[ T_m = a + (m-1)d \][/tex]
Substitute the values of [tex]\( a \)[/tex] and [tex]\( d \)[/tex]:
[tex]\[ T_m = (q + p - 1) + (m-1)(-1) \][/tex]
[tex]\[ T_m = q + p - 1 - m + 1 \][/tex]
[tex]\[ T_m = p + q - m \][/tex]

Therefore, we have shown that the [tex]\( m \)[/tex]-th term is [tex]\( p + q - m \)[/tex].

### Part (b)

Let's denote the [tex]\( m \)[/tex]-th term by [tex]\( T_m \)[/tex] and the [tex]\( n \)[/tex]-th term by [tex]\( T_n \)[/tex].

Given:
1. [tex]\( m \times T_m = n \times T_n \)[/tex]

Using the general formula again:
[tex]\[ T_m = a + (m-1)d \][/tex]
[tex]\[ T_n = a + (n-1)d \][/tex]

Substitute these into the given condition:
[tex]\[ m(a + (m-1)d) = n(a + (n-1)d) \][/tex]

Expand:
[tex]\[ ma + m(m-1)d = na + n(n-1)d \][/tex]

Rearrange to group terms involving [tex]\( a \)[/tex] and [tex]\( d \)[/tex]:
[tex]\[ ma - na = n(n-1)d - m(m-1)d \][/tex]
[tex]\[ a(m-n) = d[n(n-1) - m(m-1)] \][/tex]

Notice that:
[tex]\[ n(n-1) - m(m-1) = n^2 - n - (m^2 - m) \][/tex]
[tex]\[ = n^2 - n - m^2 + m \][/tex]

Thus, the equation becomes:
[tex]\[ a(m-n) = d(n^2 - n - m^2 + m) \][/tex]

This shows the relationship required in part (b).
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