Question

in a sequence of 21 terms the first 11 terms are in AP with common difference 2 and the last 11 terms are in GP with common ratio 2, if the middle term of the AP is equal to the middle term of the GP , then the middle term of the entire sequence ?

Answer 1

$$\begin{gathered} DearStudent, \\ Thefirst11termareinAPwithcommondifferenced=2andfirsttermisa \\ ThenthemiddletermofA.P.isthe6thterm. \\ i.e.a+\left(n-1\right)d \\ =a+\left(6-1\right)2 \\ =a+10 \\ LettheG.P.firsttermisPandcommonratior=2 \\ ThenmiddletermofG.P. \\ i.e.16thtermofsequenceor6thtermofG.P.canbewrittenas \\ =p{r}^{n-1} \\ =P.2^{6-1} \\ =P.2^5 \\ \sin cemiddletermofA.P.=middletermofG.P. \\ Therefore\left(a+10\right)=P.2^5\dots \dots \dots \dots \dots \dots \dots \left(1\right) \\ Also11thtermofA.P.=1sttermofG.P. \\ Thereforea+10.2=P \\ a+20=P\dots \dots \dots \dots \dots \dots \dots \left(2\right) \\ Solving\left(1\right)and\left(2\right)gives \\ P=\frac{-10}{31}anda=-\frac{630}{31} \\ Middletermofthesequenceisthe11thtermofA.P.or1sttermofG.P. \\ i.e.P=\frac{-10}{31} \\ Regards. \end{gathered}$$