Find the value of the middle most term(s) of the A.P. -11, -7, -3, ……., 49.
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Hint: First we need to find out the position of the last term of the given A.P. i.e. 49 using the formula $ {{a}_{n}}=a+(n-1)d $ where a is first term, d is the common difference of the A.P. and $ {{a}_{n}}=49 $. Suppose 49 is at $ {{n}^{th}} $ position in the A.P. then, if n is even then middle terms of the A.P. will be $ \dfrac{n}{2},\dfrac{n}{2}+1 $ and if n comes out to be an odd number then middle term will be $ \dfrac{n+1}{2} $. After that we will find the value of the middle term using the same formula.
Complete step-by-step answer:
First we need to find out the position of the last term of the given A.P. i.e. 49 using the formula $ {{a}_{n}}=a+(n-1)d $ , where
a is first term,
d is the common difference of the A.P. i.e. = - 7- (-11) = 4,
$ {{a}_{n}}=49 $ .
So by applying the formula and putting the required values in it, we get
49 = - 11 + (n-1) $ \times $ 4
After further simplifying, we get
$ n-1=\dfrac{49+11}{4} $
$ \begin{align}
& n-1=\dfrac{60}{4} \\
& n-1=15 \\
& n=16 \\
\end{align} $
Hence 49 is at the $ {{16}^{th}} $ place in the given A.P.
And we know that 16 is an even number so we will get two terms in the middle of the A.P. and their position will be $ \dfrac{n}{2},\dfrac{n}{2}+1 $ where n = 16,
So now we get,
$ \dfrac{16}{2},\dfrac{16}{2}+1 $ as our middle most terms i.e. $ {{8}^{th}}\,and\,{{9}^{th}} $ term of the A.P.
We will find the values of $ {{8}^{th}}\,and\,{{9}^{th}} $ terms of the A.P.
$ {{8}^{th}} $ term of the A.P. is given by,
\[\begin{align}
& {{a}_{8}}=-11+(8-1)\times 4 \\
& {{a}_{8}}=-11+7\times 4 \\
& {{a}_{8}}=-11+28 \\
& {{a}_{8}}=17 \\
\end{align}\]
Now we will find the $ {{9}^{th}} $ term of the A.P.
$ \begin{align}
& {{a}_{9}}=-11+(9-1)\times 4 \\
& {{a}_{9}}=-11+8\times 4 \\
& {{a}_{9}}=-11+32 \\
& {{a}_{9}}=21 \\
\end{align} $
Hence, value(s) of the middle most terms of the given A.P. are 17 and 21
Note: You need to remember whenever there are even numbers of terms then there will be two middle terms and when there are a middle number of terms there is only one middle term. And also remember the formula that $ {{n}^{th}} $ term of the A.P. is given by $ {{a}_{n}}=a+(n-1)d $ where a is first term and d is common difference of the A.P.
Complete step-by-step answer:
First we need to find out the position of the last term of the given A.P. i.e. 49 using the formula $ {{a}_{n}}=a+(n-1)d $ , where
a is first term,
d is the common difference of the A.P. i.e. = - 7- (-11) = 4,
$ {{a}_{n}}=49 $ .
So by applying the formula and putting the required values in it, we get
49 = - 11 + (n-1) $ \times $ 4
After further simplifying, we get
$ n-1=\dfrac{49+11}{4} $
$ \begin{align}
& n-1=\dfrac{60}{4} \\
& n-1=15 \\
& n=16 \\
\end{align} $
Hence 49 is at the $ {{16}^{th}} $ place in the given A.P.
And we know that 16 is an even number so we will get two terms in the middle of the A.P. and their position will be $ \dfrac{n}{2},\dfrac{n}{2}+1 $ where n = 16,
So now we get,
$ \dfrac{16}{2},\dfrac{16}{2}+1 $ as our middle most terms i.e. $ {{8}^{th}}\,and\,{{9}^{th}} $ term of the A.P.
We will find the values of $ {{8}^{th}}\,and\,{{9}^{th}} $ terms of the A.P.
$ {{8}^{th}} $ term of the A.P. is given by,
\[\begin{align}
& {{a}_{8}}=-11+(8-1)\times 4 \\
& {{a}_{8}}=-11+7\times 4 \\
& {{a}_{8}}=-11+28 \\
& {{a}_{8}}=17 \\
\end{align}\]
Now we will find the $ {{9}^{th}} $ term of the A.P.
$ \begin{align}
& {{a}_{9}}=-11+(9-1)\times 4 \\
& {{a}_{9}}=-11+8\times 4 \\
& {{a}_{9}}=-11+32 \\
& {{a}_{9}}=21 \\
\end{align} $
Hence, value(s) of the middle most terms of the given A.P. are 17 and 21
Note: You need to remember whenever there are even numbers of terms then there will be two middle terms and when there are a middle number of terms there is only one middle term. And also remember the formula that $ {{n}^{th}} $ term of the A.P. is given by $ {{a}_{n}}=a+(n-1)d $ where a is first term and d is common difference of the A.P.