1.

Find the 7th term of the H.P. \(\frac{2}{13}, \frac{1}{6}, \frac{2}{11}\),...........

Answer»

The reciprocals of the terms of the given H.P, i.e., \(\frac{13}{2},6,\frac{11}{2},\)......  form an A.P with first term = \(\frac{13}{2}\) and common difference = 6 - \(\frac{13}{2}\) = \(-\frac{1}{2}.\)

∴ 7th term of this A.P. = a + 6d = \(\frac{13}{2}\) + \(\bigg(-\frac{1}{2}\bigg)\) x 6 = \(\frac{7}{2}.\)

Hence 7th of the given H.P. = \(\frac{1}{\frac{7}{2}}\) = \(\frac{2}{7}.\)



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