Question

Find the middle term of an A.P with 23 terms, given its first term is 11 and common difference is 3.\[\]

Answer 1

Hint:We find the position $m$ of the middle term of the given AP using the formula $m=\left( \dfrac{n+1}{2} \right)$. We then use the formula for the ${{k}^{\text{th}}}$ term of an AP which is ${{x}_{k}}=a+\left( k-1 \right)d$ where $a$ is first term and $d$ is the common difference to find the ${{m}^{\text{th}}}$ term.\[\]

Complete step by step answer:
We know that a sequence is defined as the enumerated collection of numbers where repetitions are allowed and order of the numbers matters. It can also be expressed as a one-one map from the natural numbers set to real numbers. The members of the sequence are called terms. Mathematically, a sequence with infinite terms is written as
\[\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...\]
If the sequence has finite terms terminated by a term then we write the sequence as
\[\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...{{x}_{n}}\]
We also know that arithmetic sequence otherwise known as arithmetic progression; abbreviate d as AP is a type sequence where the difference between any two consecutive numbers is constant. If $\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...$ is an AP, then ${{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}...$ . The difference between two terms is called common difference and denoted $d$ where $d={{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}...$. Here ${{x}_{1}}$is the first term conventionally denoted by the English alphabet $a$. The ${{k}^{\text{th}}}$ term of an AP ${{x}_{n}}$ with first term $a$ and common difference $d$ is given by
\[{{x}_{k}}=a+\left( k-1 \right)d\]

We are given in the question that the AP has 23 terms. It has firs term 11 and the common difference 3. So $a=11,d=3.$
We have the total number of terms as $n=23.$ We know that if ${{m}^{\text{th}}}$ term is the middle term of the sequence with total odd number of terms $n$, then
\[m=\left( \dfrac{n+1}{2} \right)\]
So the position of the middle term of the give AP is,
\[ m=\left( \dfrac{n+1}{2} \right)=\dfrac{23+1}{2}=12 \]
So we have to find ${{12}^{\text{th}}}$of the AP. We use the formula for ${{k}^{\text{th}}}$ term of an AP and have,
\[{{x}_{m}}=a+\left( m-1 \right)d=11+\left( 12-1 \right)3=11+33=44\]
So the middle term of the AP is 44. \[\]

Note:
An arithmetic series is the expression with summation of the terms inn AP sequence. If there even number of terms we shall find two middle terms ${{m}_{1}}=\dfrac{n}{2},{{m}_{2}}=\dfrac{n}{2}+1$. The sum $S$ of first $n$ terms of an AP is $S=\dfrac{n}{2}\left\{ 2a+\left( n-1 \right)d \right\}$ and if the last term ${{a}_{n}}$ is also given then $S=\dfrac{n}{2}\left( a+{{a}_{n}} \right)$.