What is the sum of the first 12 terms of an arithmetic progression if the 3rd term is -13 and the 6th term is -4?
$$T_{3}$$ = a + 2d = -13-------(1)
$$T_{6}$$ = a + 5d = -4-------(2)
on solving (1) and (2)
d = 3 & a = -19
$$S_{n}=\frac{n}{2}[2a+(n-1)d]$$
$$S_{12}=\frac{12}{2}[2(-19)+(12-1)(3)]$$
$$S_{12}=(6)[-38+33]$$
$$S_{12}=-30$$
So the answer is option C.
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