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Mathematics

Find the sum of the two middle most terms of the A.P.

−43,−1,−23,....,413.-\dfrac{4}{3}, -1, -\dfrac{2}{3}, …., 4\dfrac{1}{3}.

AP GP

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Answer

Here, a = −43 and d =−1−(−43)=−3+43=13-\dfrac{4}{3} \text{ and d } = -1- \big(-\dfrac{4}{3}\big) = \dfrac{-3 + 4}{3} = \dfrac{1}{3}

 and l =413=133.\text{ and l } = 4\dfrac{1}{3} = \dfrac{13}{3}.

We need to find the number of terms in the A.P.

⇒an=a+(n−1)d⇒133=−43+(n−1)13⇒133+43=(n−1)3⇒173=(n−1)3\Rightarrow a_n = a + (n - 1)d \\[1em] \Rightarrow \dfrac{13}{3} = -\dfrac{4}{3} + (n - 1)\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{13}{3} + \dfrac{4}{3} = \dfrac{(n - 1)}{3} \\[1em] \Rightarrow \dfrac{17}{3} = \dfrac{(n - 1)}{3}

On multiplying both sides by 3,

⇒17=n−1⇒17+1=n⇒n=18.\Rightarrow 17 = n - 1 \\[1em] \Rightarrow 17 + 1 = n \\[1em] \Rightarrow n = 18.

Since, A.P. has 18 terms, therefore 9th and 10th terms are two middle most terms.

a9 = a + (9 - 1)d = a + 8d        (Eq 1)

a10 = a + (10 - 1)d = a + 9d     (Eq 2)

Adding Eq 1 and Eq 2,

a9 + a10 = a + 8d + a + 9d = 2a + 17d.

Hence, the sum of middle most terms

=2a+17d=2×−43+17×13=−83+173=−8+173=93=3.= 2a + 17d \\[0.5em] = 2 \times -\dfrac{4}{3} + 17 \times \dfrac{1}{3} \\[0.5em] = -\dfrac{8}{3} + \dfrac{17}{3} \\[0.5em] = \dfrac{-8 + 17}{3} \\[0.5em] = \dfrac{9}{3} \\[0.5em] = 3.

Hence, the sum of the two middle most terms of the A.P. is equal to 3.

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