What is the sum of the first 11 terms of an arithmetic progression if the 3rd term is -1 and the 8th term is 19?
$$T_{3}$$ = a + 2d = -1-------(1)
$$T_{8}$$ = a + 7d = 19-------(2)
on solving (1) AND (2)
d = 4 & a = -9
$$S_{n}=\frac{n}{2}[2a+(n-1)d]$$
$$S_{11}=\frac{11}{2}[2(-9)+(11-1)(4)]$$
$$S_{11}=\frac{11}{2}[(-18)+(40)]$$
$$S_{11}=\frac{11}{2}[22]$$
$$S_{11}=121$$
So the answer is option B.
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