Question & Answer
QUESTION

Find the middle terms of the A.P. \[7,13,19,............................,241.\]
A. \[411,127\]
B. \[121,227\]
C. \[121,127\]
D. \[321,127\]

ANSWER Verified Verified
Hint: First of all, find the total number of terms by using the first term, last term and common difference of the series. Then find the middle terms of the given series by using the formula \[{t_n} = a + \left( {n - 1} \right)d\]. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:

Given series of A.P. is \[7,13,19,............................,241.\]

We know that for a series of A.P. the last term is given by \[l = a + \left( {n - 1} \right)d\]

where ‘\[a\]’ is the first term, ‘\[n\]’ is the number of terms, ‘\[d\]’ is the common difference of the given series.

In the giver series clearly the first term is 7 and the last term is 241.

Also, we know the common difference is given the difference of the first and second terms i.e., \[d = 13 - 7 = 6\].

By using the above data, we have

\[

   \Rightarrow l = a + \left( {n - 1} \right)d \\

   \Rightarrow 241 = 7 + \left( {n - 1} \right)6 \\

   \Rightarrow 241 - 7 = 6n - 6 \\

   \Rightarrow 241 - 7 + 6 = 6n \\

   \Rightarrow 240 = 6n \\

  \therefore n = \dfrac{{240}}{6} = 40 \\

\]

Hence the number of terms in the series is 40

As the number of terms (\[n\]) is an even integer, that middle terms are given by \[{\left( {\dfrac{n}{2}} \right)^{th}},{\left( {\dfrac{n}{2} + 1} \right)^{th}}\] terms.

We know that the \[{n^{th}}\] term of a series in A.P is given by \[{t_n} = a + \left( {n - 1} \right)d\]

So, the \[{\left( {\dfrac{n}{2}} \right)^{th}}\]terms i.e., 20th term is given by

\[

   \Rightarrow {t_{20}} = 7 + \left( {20 - 1} \right)6 \\

   \Rightarrow {t_{20}} = 7 + 19 \times 6 \\

  \therefore {t_{20}} = 121 \\

\]

And the \[{\left( {\dfrac{n}{2} + 1} \right)^{th}}\] terms i.e., 21th term is given by

\[

   \Rightarrow {t_{21}} = 7 + \left( {21 - 1} \right)6 \\

   \Rightarrow {t_{21}} = 7 + 20 \times 6 \\

  \therefore {t_{21}} = 127 \\

\]

So, the middle terms of the given A.P. are 121 and127.

Thus, the correct option is C. \[121,127\]

Note: For a series of A.P. the last term is given by \[l = a + \left( {n - 1} \right)d\] where ‘\[a\]’ is the first term, ‘\[n\]’ is the number of terms, ‘\[d\]’ is the common difference of the given series. If the number of terms (\[n\]) is an even integer, that middle terms are given by \[{\left( {\dfrac{n}{2}} \right)^{th}},{\left( {\dfrac{n}{2} + 1} \right)^{th}}\] terms.