Question

How many terms are there in the ap whose first and eighth terms are -16 and -2 respectively and sum of their terms is 18

Answer 1

Dear student,

According to the question. We know that,

Let d be the common difference.

First term (a) = -16

Eighth term (a₈) = -2

a_n = a + (n - 1)d

or, a₈ = -16 + (8-1)d

or, -2 = -16 + 7d

​or, 14 = 7d

So, d = 2

Also, we know that the sum of first n terms of an AP is : ${\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\hspace{0.17em}\left(\mathrm{n}-1\right)\mathrm{d}\right]$

Let n be the number of terms in the given AP


$$\begin{gathered} {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right] \\ \Rightarrow 18=\frac{\mathrm{n}}{2}\left[2\left(-16\right)+\hspace{0.17em}\left(\mathrm{n}-1\right)2\right] \\ \Rightarrow 18=\frac{\mathrm{n}}{2}\left[-32+\left(\mathrm{n}-1\right)2\right] \\ \Rightarrow 18=-16\mathrm{n}+\left(\mathrm{n}-1\right)\mathrm{n} \\ \Rightarrow 18=-16\mathrm{n}+\hspace{0.17em}{\mathrm{n}}^2-\mathrm{n} \\ \Rightarrow 18=-17\mathrm{n}+\hspace{0.17em}{\mathrm{n}}^2 \\ \Rightarrow {\mathrm{n}}^2-17\mathrm{n}-18=0 \\ \Rightarrow {\mathrm{n}}^2-18\mathrm{n}+\mathrm{n}-18=0 \\ \Rightarrow \mathrm{n}\left(\mathrm{n}-18\right)+1\left(\mathrm{n}-18\right)=0 \\ \Rightarrow \left(\mathrm{n}-18\right)\left(\mathrm{n}+1\right)=0 \\ \Rightarrow \mathrm{n}=18\mathrm{or}\mathrm{n}=-1\left(\mathrm{rejected}\right) \end{gathered}$$

Hence, there will be 18 terms in the given AP.