Question:
The 11 th term of an AP $-5, \frac{-5}{2}, 0, \frac{5}{2} \ldots$
(a)-20
(b) 20
(c) $-30$
(d) 30
Solution:
Given AP, $5, \frac{-5}{2}, 0, \frac{5}{2}$
Here, $a=-5, d=\frac{-5}{2}+5=\frac{5}{2}$
$\therefore \quad a_{11}=a+(11-1) d \quad\left[\because a_{n}=a+(n-1) d\right]$
$=-5+(10) \times \frac{5}{2}$
$=-5+25=20$
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