Question

the sum of 7th and third terms of an AP is 6 and their product is 8.find the sum of first 20 terms of the AP.

Answer 1

$$\begin{gathered} a+6d=7thtermoftheA.P. \\ a+2d=3rdtermoftheA.P. \\ Accordingtothequestion: \\ a+6d+a+2d=6 \\ 2a+8d=6 \\ a+4d=3 \\ a=3-4d....\left(1\right) \\ And,(a+6d)(a+2d)=8 \\ \left(3-4d+6d\right)\left(3-4d+2d\right)=8 \\ \left(3+2d\right)\left(3-2d\right)=8 \\ \left(9-4d^2\right)=8 \\ -4d^2=8-9 \\ {d}^2=\frac{1}{4} \\ d=\pm \frac{1}{2} \\ Putthevalueofdintheequation\left(1\right),weget, \\ a=3-4d \\ a=3-4\left(\pm \frac{1}{2}\right) \\ a=1or,5 \\ Now,S_n=\frac{n}{2}\left\{2a+\left((n-1)d\right)\right\} \\ \Rightarrow S_{20}=\frac{20}{2}\left\{2\left(1\right)+\left(20-1\right)\frac{1}{2}\right\}(Whend=\frac{1}{2}) \\ \Rightarrow S_{20}=10\left\{2+\left(20-1\right)\frac{1}{2}\right\} \\ \Rightarrow S_{20}=20+\frac{190}{2}=\frac{230}{2}=115 \\ And, \\ \Rightarrow S_{20}=\frac{20}{2}\left\{2\left(1\right)+\left(20-1\right)(-\frac{1}{2})\right\}(Whend=-\frac{1}{2}) \\ \Rightarrow S_{20}=10\left\{2+\left(20-1\right)(-\frac{1}{2})\right\} \\ \Rightarrow S_{20}=20-\frac{190}{2}=\frac{-150}{2}=-75 \end{gathered}$$