Question 6

If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term

Answer

Let the sum of n terms of the given A.P. be 116.

\begin{align} S_n=\frac{n}{2}\left[2a + (n-1)d\right] \end{align}

Here, a = 25 and d = 22 – 25 = – 3

\begin{align} \therefore S_n=\frac{n}{2}\left[2 × 25 + (n-1)(-3)\right] \end{align}

\begin{align} ⇒ 116=\frac{n}{2}\left[50 -3n +3\right] \end{align}

\begin{align} ⇒ 232=n(53-3n)=53n -3n^2 \end{align}

\begin{align} ⇒ 3n^2 -53n + 232 =0\end{align}

\begin{align} ⇒ 3n^2 -24n -29n + 232 =0\end{align}

\begin{align} ⇒ 3n(n-8) -29(n-8) =0\end{align}

\begin{align} ⇒ (n-8)(3n-29) =0\end{align}

\begin{align} ⇒ n=8 \;or\;n=\frac{29}{3} \end{align}

However, n cannot be equal to \begin{align} \frac{29}{3}. \end{align} Therefore, n = 8

\begin{align} \therefore a_8 = Last \; Term = a + (n-1)d= 25 + (8-1)(-3) \end{align}

\begin{align} = 25 + (7)(-3)=25-21 \end{align}

\begin{align} = 4 \end{align}

Thus, the last term of the A.P. is 4.

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