Find the sum of the sequence -1, -5/6, -2/3, -1/2, 10/3

Question

Find the sum of the sequence

−1\(\frac{−5}{6}\), \(\frac{−2}{3}\), \(\frac{−1}{2}\) …,\(\frac{10}{3}\)

Answer 1

Given sequence in A.P.

−1 \(\frac{−5}{6}\), \(\frac{−2}{3}\), \(\frac{−1}{2}\) … ,\(\frac{10}{3}\)

where a = −1, d = \(\frac{−5}{6}\) + 1 = \(\frac{1}{6}\) and I = \(\frac{10}{3}\)

Using formula I = a + (n − 1)d, we get

\(\frac{10}{3}\) = −1 + (n − 1) \(\frac{1}{6}\)

⇒ \(\frac{10}{3}\) + 1 = \(\frac{n}{6}\) − \(\frac{1}{6}\)

⇒ \(\frac{13}{3}\) + \(\frac{1}{6}\) = \(\frac{n}{6}\)

⇒ \(\frac{27}{3}\) = \(\frac{n}{6}\)

⇒ n = 27

Now, sum of Nh terms of an A.P. is

S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

∴ S₁₅ = \(\frac{27}{2}\)\(\big[2(−1) + (26)\frac{1}{6}\big]\)

= \(\frac{27}{2}\)\(\big[−2 +\frac{26}{6} \big]\)

= \(\frac{27}{2}\)\(\big[\frac{14}{6} \big]\)

= \(\frac{63}{2}\)