Additional Mathematics 0606 / Series / 3.633.63: Log/exponential equations; AP and dual GPs0606/13/O/N/25 — Question 9 · 14 marksMark as done · Save for later · Show all solutions(a)Solve 2log5(5x−2)−2log25x=12\log_{5}(5x-2)-2\log_{25}x=12log5(5x−2)−2log25x=1.[5]▸ Answer(b)Solve e3y−7+4e−3=5e1−3y\mathrm{e}^{3y-7}+4\mathrm{e}^{-3}=5\mathrm{e}^{1-3y}e3y−7+4e−3=5e1−3y.[6]▸ Answer(a)An AP has S20=3S10S_{20}=3S_{10}S20=3S10. Find aaa in terms of ddd.[3]▸ Answer(b)GP AAA has ratio rrr with ∣r∣<1|r|<1∣r∣<1. GP BBB has terms bk=a2kb_{k}=a_{2k}bk=a2k. Find SBSA\dfrac{S_{B}}{S_{A}}SASB in terms of rrr.[5]▸ Answer▸ Official mark scheme← 3.62: GP with S2=9S_{2}=9S2=9 and S∞=25S_{\infty}=25S∞=253.65: AP terms forming a geometric progression →
