Question

Solve: (i) 25 + 22 + 19 + 16 + ... + x = 115
(ii) 1 + 4 + 7 + 10 + ... + x = 590.

Answer 1

(i) 25 + 22 + 19 + 16 + ... + x = 115
Here, a = 25, d = $-$3, $S_n=115$
$$\begin{gathered} \mathrm{We}\mathrm{know}: \\ {S}_n=\frac{n}{2}\left[2a+(n-1)d\right] \\ \Rightarrow 115=\frac{n}{2}\left[2\times 25+(n-1)\times \left(-3\right)\right] \\ \Rightarrow 115\times 2=n\left[50-3n+3\right] \\ \Rightarrow 230=n\left(53-3n\right) \\ \Rightarrow 230=53n-3n^2 \\ \Rightarrow 3n^2-53n+230=0 \\ \mathrm{By}\mathrm{quadratic}\mathrm{formula}: \\ n=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \mathrm{Substituting}a=3,b=-53\mathrm{and}c=230,\mathrm{we}\mathrm{get}: \\ n=\frac{53\pm \sqrt{{\left(53\right)}^2-4\times 3\times 230}}{2\times 3}=\frac{46}{6},10 \\ \Rightarrow n=10,asn\ne \frac{46}{6} \\ \therefore a_n=a+(n-1)d \\ \Rightarrow x=25+(10-1)(-3) \\ \Rightarrow x=25-27=-2 \end{gathered}$$

(ii) 1 + 4 + 7 + 10 + ... + x = 590
Here, a = 1, d = 3,
$$\begin{gathered} \mathrm{We}\mathrm{know}: \\ {S}_n=\frac{n}{2}\left[2a+(n-1)d\right] \\ \Rightarrow 590=\frac{n}{2}\left[2\times 1+(n-1)\times \left(3\right)\right] \\ \Rightarrow 590\times 2=n\left[2+3n-3\right] \\ \Rightarrow 1180=n\left(3n-1\right) \\ \Rightarrow 1180=3n^2-n \\ \Rightarrow 3n^2-n-1180=0 \\ \mathrm{By}\mathrm{quadratic}\mathrm{formula}: \\ n=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \mathrm{Substituting}a=3,b=-1\mathrm{and}c=-1180,\mathrm{we}\mathrm{get}: \\ \Rightarrow n=\frac{1\pm \sqrt{{\left(1\right)}^2+4\times 3\times 1180}}{2\times 3}=\frac{-118}{6},20 \\ \Rightarrow n=20,asn\ne \frac{-118}{6} \\ \therefore a_n=x=a+(n-1)d \\ \Rightarrow x=1+(20-1)\left(3\right) \\ \Rightarrow x=1+60-3=58 \end{gathered}$$