Question

The sum of 11 terms of an A.P, whose middle term is 30, is
$\left( a \right)$ 320
$\left( b \right)$ 330
$\left( c \right)$ 340
$\left( d \right)$ 350

Answer 1

Hint: In this question use the concept that in A.P any general term is given as, [a + (k– 1) d], where a is first term, d is common difference and k is the number of particular term, and sum of the A.P is given as, ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$,where n is the number of terms.

Complete step-by-step answer:
Given data:
There are 11 terms in an A.P, so n = 11.
It is also given that the value of the middle term of an A.P is 30.
Now the number of terms is 11 which is odd.
So the middle is calculated as, $\dfrac{{n + 1}}{2} = \dfrac{{11 + 1}}{2} = \dfrac{{12}}{2} = 6$.
Now as we know that in A.P any general term is given as, [a + (k– 1) d], where a is first term, d is a common difference and k is the number of particular terms.
So ${6^{th}}$ term of an A.P is,
Therefore, k = 6
$ \Rightarrow 30 = a + \left( {6 - 1} \right)d$
$ \Rightarrow 30 = a + 5d$..................... (1)
Now as we know that the sum of an A.P series is given as,
${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where symbols have their usual meanings.
Now substitute the value of n we have,
$ \Rightarrow {S_n} = \dfrac{{11}}{2}\left( {2a + \left( {11 - 1} \right)d} \right)$
$ \Rightarrow {S_n} = \dfrac{{11}}{2}\left( {2a + 10d} \right)$
$ \Rightarrow {S_n} = 11\left( {a + 5d} \right)$
Now substitute the value of (a + 5d) from equation (1) we have,
$ \Rightarrow {S_n} = 11\left( {30} \right) = 330$
So this is the required sum of 11 term A.

So, the correct answer is “Option b”.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall all the basic formulas of A.P which are all stated above them simplify just substitute the values as above substituted we will get the required answer.