Question

prove that if the number of terms of an A.P is odd then the middle term is the A P between the first and last term???​

Answer 1

Prove that if the number of terms of an A.P is odd then the middle term is the A.M between the first and last term???​

Answer:

Proved that when number of terms of an A.P is odd then the middle term is the A .M between the first and last term

Solution:

To prove: If the number of terms of an A.P is odd then the middle term is the A .M between the first and last term

Proof:

Let the arithmetic progression be:

a, a + d, a + 2d, ........., (2n + 1) terms

As the number of terms are odd, the nth term is (2n + 1)

The middle term of (2n + 1)th term is:

$\rightarrow \frac{(2n + 1 + 1)}{2} = \frac{2n + 2}{2}= (n + 1)$

Hence middle term is (n + 1)th term

The formula for nth term of arithmetic progression is:

$a_n = a + (n - 1)d$  ------- (i)

Where n is the terms location and "d" is the common difference between terms

Thus middle term is: (n + 1)th term

From (i)

$\text{ middle term } = a + (n + 1 - 1)d = a + nd$  ----- (ii)

Arithmetic mean between first and last term:

The nth term is (2n + 1)

Hence nth term (last term) is:

From (i)

$\rightarrow a + (2n + 1 - 1)d = a + 2nd$

So Arithmetic mean between 1st & last term is:

First term of A.P = a

Arithmetic mean between first and last term = (first term + last term)/2

$\rightarrow \frac{a + a + 2nd}{2}\\\\\rightarrow \frac{2a + 2nd}{2}\\\\\rightarrow a + nd$  ----- (iii)

Thus from steps (ii) and (iii)

Middle term = AM of 1st and last terms [Proved]

Thus proved that when number of terms of an A.P is odd then the middle term is the A .M between the first and last term