Question 113
The 288th term of the sequence a, b, b, c, c, c, d, d, d, d… is
Solution
Here, $$\ \frac{\ n\left(n-1\right)}{2}$$(not inclusive) term to $$\ \frac{\ n\left(n+1\right)}{2}$$(inclusive) term is the nth place alphabet. (1st place alphabet is a, 2nd place alphabet is b, and so on.)
=> $$\ \frac{\ n\left(n-1\right)}{2}$$ < 288 < $$\ \frac{\ n\left(n+1\right)}{2}$$
=> n(n-1) < 576 < n(n+1)
n = 24
At 24th place, x is the alphabet.
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