Correct Answer: 4796

Solution : Given: $98^2-97^2+96^2-95^2+94^2-93^2+\ldots \ldots 12^2-11^2$
$=(98+97)(98-97)+(96+95)(96-95)+(94+93)(94-93).......(12+11)(12-11)$
$=195+191+187+......+23$
Here $a=195$, $d=191 - 195 = -4$
$T_n = a + (n-1)d$
⇒ $23 = 195 + (n-1)(-4)$
⇒ $n-1=\frac{23-195}{-4}$
⇒ $n-1=\frac{-172}{-4}$
⇒ $n-1=43$
⇒ $n=44$
Sum of the series = $\frac{n}{2}(a_1 + a_n)$, where $n$ is the number of terms, $a_1$ is the first term, and $a_n$ is the last term.
Here, $n = 44$ (from 195 to 23)
$a_1 = 195$
$a_n = 23$
So, the sum of the series = $\frac{44}{2}(195 + 23) = 22 \times 218 = 4796$
Hence, the correct answer is 4796.