పరిష్కారం - జ్యామితీ నిర్ణయాలు

శ్రేణికి సామాన్య రూపాన్ని కనుగొనడానికి, మొదటి పదాన్ని: $$a=99$$ మరియు పునరావృత్తి అనుపాతాన్ని: $$r=0.020202020202020204$$ జామితీయ శ్రేణుల సూత్రానికి ప్లగ్ చేయండి:

$$a_1=99$$

$$a_2=a_1·r^{n-1}=99\cdot {0.020202020202020204}^{2-1}=99\cdot {0.020202020202020204}^1=99\cdot 0.020202020202020204=2$$

$$a_3=a_1·r^{n-1}=99\cdot {0.020202020202020204}^{3-1}=99\cdot {0.020202020202020204}^2=99\cdot 0.0004081216202428324=0.04040404040404041$$

$$a_4=a_1·r^{n-1}=99\cdot {0.020202020202020204}^{4-1}=99\cdot {0.020202020202020204}^3=99\cdot 8.244881217026918E-06=0.0008162432404856649$$

$$a_5=a_1·r^{n-1}=99\cdot {0.020202020202020204}^{5-1}=99\cdot {0.020202020202020204}^4=99\cdot 1.6656325690963473E-07=1.6489762434053837E-05$$

$$a_6=a_1·r^{n-1}=99\cdot {0.020202020202020204}^{6-1}=99\cdot {0.020202020202020204}^5=99\cdot 3.364914281002722E-09=3.331265138192695E-07$$

$$a_7=a_1·r^{n-1}=99\cdot {0.020202020202020204}^{7-1}=99\cdot {0.020202020202020204}^6=99\cdot 6.797806628288329E-11=6.729828562005445E-09$$

$$a_8=a_1·r^{n-1}=99\cdot {0.020202020202020204}^{8-1}=99\cdot {0.020202020202020204}^7=99\cdot 1.3732942683410765E-12=1.3595613256576657E-10$$

$$a_9=a_1·r^{n-1}=99\cdot {0.020202020202020204}^{9-1}=99\cdot {0.020202020202020204}^8=99\cdot 2.774331855234498E-14=2.7465885366821533E-12$$

$$a_{10}=a_1·r^{n-1}=99\cdot {0.020202020202020204}^{10-1}=99\cdot {0.020202020202020204}^9=99\cdot 5.604710818655553E-16=5.5486637104689975E-14$$