Question

Find the sum of all the 11 terms of an A.P. whose middle terms is 30.

Answer 1

For solving this question, we will use the concept of ${{\text{n}}^{\text{th}}}$ term of an A . P whose first term is a and difference is d and the sum of n terms of an A..P. We will first evaluate the which is the place of term 30 in an A.P. And, then we will substitute the values of a and d term in a formula of sum of an A.P.

Complete step-by-step answer :
In question it is provided that the total number of terms in an A.P is equals to 11 and the value of the middle most term in an A.P is equals to 30.
So, firstly we need to calculate at which term 30 occurs.
Now, here n = 11 which is odd. So, middle term of series when n is odd $={{\dfrac{(n+1)}{2}}^{th}}term$
So, middle term for A.P of 11 term $={{\dfrac{(11+1)}{2}}^{th}}term$
On solving we get,
${{6}^{th}}term=30$
Now, for any term at n place in A.P van be evaluated by formula ${{a}_{n}}=a+(n-1)\cdot d$ , where a = first term , n = place of term , d = common difference and ${{a}_{n}}=$ term at ${{n}^{th}}$ place.
So, ${{a}_{6}}=a+(6-1)\cdot d$
Or $30-5d=a$
Now, summation of n terms of an A.P is equals to ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Substituting values of a , and n in ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ , we get
${{S}_{11}}=\dfrac{11}{2}\left[ 2a+\left( 11-1 \right)d \right]$…….( i )
Substituting $30-5d=a$ in equation ( i ), we get
${{S}_{11}}=\dfrac{11}{2}\left[ 60-10d+10d \right]$
On simplifying, we get
${{S}_{11}}=\dfrac{11}{2}\left( 60 \right)$
On solving we get
${{S}_{11}}=330$
Hence, the sum of all 11 terms of an A.P is equals to 330.

Note : As we see on substituting the values of a, b and c, we get a very complex equation so try to avoid calculation error as it may affect the answer and change the value. Re – arranging of the terms should be done in such a way that calculation becomes a bit easy and terms get cancelled out. Also , sometimes it's not necessary to solve the linear equation, we can substitute the linear equation in another complex equation to make it into simpler form and evaluate the answer.