Question 4

How many terms of the A.P. \begin{align} -6, -\frac{11}{2}, -5, ... \end{align} are needed to give the sum –25?

Answer

Let the sum of n terms of the given A.P. be –25.

It is known that, \begin{align} S_n = \frac {n}{2}\left[2a + (n -1)d\right]\end{align}, where n = number of terms, a = first term, and d = common difference

Here, a = –6

\begin{align} d = -\frac{11}{2} + 6 = \frac{-11 + 12}{2} = \frac{1}{2}\end{align}

Therefore, we obtain

\begin{align} -25 = \frac {n}{2}\left[2 × (-6) + (n -1)×\frac{1}{2}\right]\end{align}

\begin{align} => -50 = n\left[-12 + \frac{n}{2} - \frac{1}{2}\right]\end{align}

\begin{align} => -50 = n\left[-\frac{25}{2} + \frac{n}{2}\right]\end{align}

\begin{align} => -100 = n\left(-25 + n\right)\end{align}

\begin{align} => n^2 - 25n + 100 = 0\end{align}

\begin{align} => n^2 - 5n -20n + 100 = 0\end{align}

\begin{align} => n(n - 5)- 20(n - 5) = 0\end{align}

\begin{align} => n = 20 \; or\; 5\end{align}

How many terms of the A.P. \begin{align} -6, -\frac{11}{2}, -5, ... \end{align} are needed to give the sum –25?

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