Question

the sum of 11 terms of an ap whose middle term is 30.​

Answer 1

For such questions, we use a formula.

$S_n=n\left(T_{\frac{n+1}{2}}\right)$

Means, to find the sum of 'n' terms of an AP if 'n' (no. of terms) and middle term are given, multiply 'n' and middle term! That's all!

Here the middle term is [(n + 1) / 2]th term. We call this as median.

Here, given that,

→  No. of terms = 11

→  Middle term = 30.

Here middle term is 6th term.

(11 + 1) / 2 = 12 / 2 = 6

Hence, the sum is "330".

11 × 30 = 330

Now, proof of the above applied formula is given below.

Consider an AP  a, a + d, a + 2d, a + 3d,......, a + (n - 1)d.

Here first term is a, common difference is d, and nth term is a + (n - 1)d.

And we know about the sum.

$S_n=\dfrac{n}{2}\bigg[2a+(n-1)d\bigg]$

Now we find [(n + 1) / 2]th term.

$T_{\frac{n+1}{2}}=a+\left(\dfrac{n+1}{2}-1\right)d\\ \\ \\ T_{\frac{n+1}{2}}=a+\left(\dfrac{n+1-2}{2}\right)d\\ \\ \\ T_{\frac{n+1}{2}}=a+\left(\dfrac{n-1}{2}\right)d\\ \\ \\ T_{\frac{n+1}{2}}=\dfrac{2a+(n-1)d}{2}$

Now we multiply 'n' to both sides.

$n\left(T_{\frac{n+1}{2}}\right)=n\cdot \dfrac{2a+(n-1)d}{2}\\ \\ \\ n\left(T_{\frac{n+1}{2}}\right)=\dfrac{n(2a+(n-1)d)}{2}\\ \\ \\ n\left(T_{\frac{n+1}{2}}\right)=\dfrac{n}{2}\bigg[2a+(n-1)d\bigg]$

Hence Proved!