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Consider the A.P. $2,5,8,11,.......,302$ .Show that twice of the middle term of the above A.P. is equal to the sum of its first and last term.

Last updated date: 29th Mar 2023
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Hint: Use ${n^{th}}$ term formula ${a_n} = a + \left( {n - 1} \right)d$

It is given $2,5,8,11,.......,302$ is an A.P. with first term $a = 2$ and common difference $d = \left( {II} \right)term - \left( I \right)term = 5 - 2 = 3$ .Let there be $n$ terms in the given A.P.
Then, ${n^{th}}$ term $\left( {{a_n}} \right) = 302$
Formula of ${n^{th}}$term is ${a_n} = a + \left( {n - 1} \right)d$
$\Rightarrow a + \left( {n - 1} \right)d = 302 \\ \Rightarrow 2 + 3\left( {n - 1} \right) = 302 \\ \Rightarrow 2 + 3n - 3 = 302 \\ \Rightarrow 3n = 303 \\ \Rightarrow n = 101 \\ \\$
Clearly, $n$ is odd. Therefore , ${\left( {\dfrac{{n + 1}}{2}} \right)^{th}}$ i.e. ${51^{st}}$ term is the middle term.
Now, middle term $= {a_{51}} = a + 50d = 2 + 50 \times 3 = 152$
Sum of first term and last term $= 2 + 302 = 304$
Twice of middle term $= 2 \times 152 = 304$
So, Twice of the middle term of the above A.P. is equal to the sum of its first and last term.

Note: Whenever we come across these types of problems first we have to find the first term and common difference of given A.P., then using ${n^{th}}$ term formula find the number of terms in an A.P. If number of terms is odd then for middle term use ${\left( {\dfrac{{n + 1}}{2}} \right)^{th}}$
After finding the middle term again use the ${n^{th}}$ term formula to find the value of the middle term.