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The 9th term of an arithmetic progression (AP) is 0. Show that the 29th term of the AP is two times the 19th term of the AP.

[Hint: Obtain [tex]a = -8d[/tex] from [tex]a_9 = 0[/tex], and then find the ratio [tex]\frac{a_{29}}{a_{19}}[/tex].]

Answer :

Sure! Let's solve the problem step-by-step.

### Step 1: Establish the Formulas and Given Information
For an Arithmetic Progression (AP):
- The [tex]\( n \)[/tex]-th term ([tex]\( a_n \)[/tex]) is given by:

[tex]\[
a_n = a + (n-1)d
\][/tex]

where [tex]\( a \)[/tex] is the first term, and [tex]\( d \)[/tex] is the common difference.

- We are given that the 9th term of the AP is [tex]\( 0 \)[/tex]:

[tex]\[
a + 8d = 0
\][/tex]

### Step 2: Find the First Term [tex]\( a \)[/tex]
From the given [tex]\( a + 8d = 0 \)[/tex], we can solve for [tex]\( a \)[/tex]:

[tex]\[
a = -8d
\][/tex]

### Step 3: Express the 29th and 19th Terms of the AP
Using the formula for the [tex]\( n \)[/tex]-th term in an AP:

- The 29th term ([tex]\( a_{29} \)[/tex]) is:

[tex]\[
a_{29} = a + 28d
\][/tex]

- The 19th term ([tex]\( a_{19} \)[/tex]) is:

[tex]\[
a_{19} = a + 18d
\][/tex]

### Step 4: Substitute [tex]\( a = -8d \)[/tex] into the Formulas
Now plug [tex]\( a = -8d \)[/tex] into the expressions for [tex]\( a_{29} \)[/tex] and [tex]\( a_{19} \)[/tex]:

- For [tex]\( a_{29} \)[/tex]:

[tex]\[
a_{29} = -8d + 28d
\][/tex]

Simplify:

[tex]\[
a_{29} = 20d
\][/tex]

- For [tex]\( a_{19} \)[/tex]:

[tex]\[
a_{19} = -8d + 18d
\][/tex]

Simplify:

[tex]\[
a_{19} = 10d
\][/tex]

### Step 5: Find the Ratio [tex]\(\frac{a_{29}}{a_{19}}\)[/tex]
Now calculate the ratio of the 29th term to the 19th term:

[tex]\[
\frac{a_{29}}{a_{19}} = \frac{20d}{10d}
\][/tex]

Simplify the ratio:

[tex]\[
\frac{a_{29}}{a_{19}} = 2
\][/tex]

### Conclusion
We have shown that the 29th term of the AP is indeed two times the 19th term:

[tex]\[
a_{29} = 2 \times a_{19}
\][/tex]

Thus, the relationship between the 29th term and the 19th term is confirmed as required.

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