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Mathematics

Refer the given sequence 23, 211221\dfrac{1}{2}, 20, ….

(a) Find the general term of the given sequence.

(b) Which term is the last positive term in the sequence.

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Answer

(a) Given,

23, 211221\dfrac{1}{2}, 20, …. is A.P.

a = 23

d = 2112−23=−112=−3221\dfrac{1}{2} - 23 = -1\dfrac{1}{2} = -\dfrac{3}{2}.

We know that,

an = a + (n - 1)d

⇒an=23−(32)(n−1)⇒an=23−(3n−32)⇒an=(46−3n+32)⇒an=(49−3n2)\Rightarrow an = 23 - \Big(\dfrac{3}{2}\Big)(n - 1) \\[1em] \Rightarrow an = 23 - \Big(\dfrac{3n - 3}{2}\Big) \\[1em] \Rightarrow an = \Big(\dfrac{46 - 3n + 3}{2}\Big) \\[1em] \Rightarrow an = \Big(\dfrac{49 - 3n}{2}\Big)

Hence, an = (49−3n2)\Big(\dfrac{49 - 3n}{2}\Big) .

(b) The last positive term occurs when :

⇒ an > 0

⇒ 49−3n2>0\dfrac{49 - 3n}{2} \gt 0

⇒ 49 - 3n > 0

⇒ 3n < 49

⇒ n < 493\dfrac{49}{3}

⇒ n < 161316\dfrac{1}{3}

Thus, 16th term will be the last positive term of the sequence.

⇒ a16 = (49−3×162)=49−482=12\Big(\dfrac{49 - 3 \times 16}{2}\Big) = \dfrac{49 - 48}{2} = \dfrac{1}{2} = 0.5

Hence, the 16th term is the last positive term.

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