Question 39

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?

Solution

$$T_{3}$$ = 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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