Question

What is the ${{25}^{th}}$ term of an arithmetic sequence where ${{a}_{1}}=8$ and ${{a}_{9}}=48$?

Answer 1

For solving this question you should know about the arithmetic sequence. In the arithmetic sequence we can find a term like arithmetic progression. We have to only find the first term and the sequence between two continuous terms and the terms place which we have to find.

Complete step by step answer:
In our question we have to find the ${{25}^{th}}$ term of an arithmetic sequence where the first term of the sequence is ${{a}_{1}}=8$ and the ${{9}^{th}}$ term of the sequence is 48.
We can calculate the ${{n}^{th}}$ term of a sequence by this very easily and this is the best method for calculating the ${{n}^{th}}$ value of sequence.
So, we can write ${{a}_{1}}=8$, ${{a}_{9}}=48$.
Now, we have to find the sequence between two continuous terms.
So, the formula for finding any member of sequence at position “n”.
${{a}_{n}}={{a}_{1}}+\left( n-1 \right)\times d$
Here, ${{A}_{n}}$ is the ${{n}^{th}}$ term which we have to find.
By the given value we can write

& d=\dfrac{\left( {{a}_{9}}-{{a}_{1}} \right)}{\left( 9-1 \right)} \\\ & d=\dfrac{48-8}{8} \\\ & d=\dfrac{40}{8}=5 \\\ \end{aligned}$$ So, the sequence between the two continuous values or terms is 5. Now, we have to calculate the $${{25}^{th}}$$ value of the sequence. So, $$n=25$$. The $${{25}^{th}}$$ value of the sequence $${{a}_{25}}$$ defined by the formula is: $$\begin{aligned} & {{a}_{25}}={{a}_{1}}+\left( 25-1 \right)\times 5 \\\ & {{a}_{25}}=8+120 \\\ & {{a}_{25}}=128 \\\ \end{aligned}$$ **So, the $${{25}^{th}}$$ term of the sequence is 128.** **Note:** The arithmetic sequence is only used if we are calculating any term for a sequence. It is not used in the calculation of Arithmetic progression (A.P.). Because both are different terms so they can not be calculated by the same formula. It is always in a fixed sequence and always in fixed values.