పరిష్కారం - జ్యామితీ నిర్ణయాలు

శ్రేణికి సామాన్య రూపాన్ని కనుగొనడానికి, మొదటి పదాన్ని: $$a=-20$$ మరియు పునరావృత్తి అనుపాతాన్ని: $$r=-0.2$$ జామితీయ శ్రేణుల సూత్రానికి ప్లగ్ చేయండి:

$$a_1=-20$$

$$a_2=a_1·r^{n-1}=-20\cdot -{0.2}^{2-1}=-20\cdot -{0.2}^1=-20\cdot -0.2=4$$

$$a_3=a_1·r^{n-1}=-20\cdot -{0.2}^{3-1}=-20\cdot -{0.2}^2=-20\cdot 0.04000000000000001=-0.8000000000000002$$

$$a_4=a_1·r^{n-1}=-20\cdot -{0.2}^{4-1}=-20\cdot -{0.2}^3=-20\cdot -0.008000000000000002=0.16000000000000003$$

$$a_5=a_1·r^{n-1}=-20\cdot -{0.2}^{5-1}=-20\cdot -{0.2}^4=-20\cdot 0.0016000000000000003=-0.03200000000000001$$

$$a_6=a_1·r^{n-1}=-20\cdot -{0.2}^{6-1}=-20\cdot -{0.2}^5=-20\cdot -0.0003200000000000001=0.006400000000000001$$

$$a_7=a_1·r^{n-1}=-20\cdot -{0.2}^{7-1}=-20\cdot -{0.2}^6=-20\cdot 6.400000000000002E-05=-0.0012800000000000005$$

$$a_8=a_1·r^{n-1}=-20\cdot -{0.2}^{8-1}=-20\cdot -{0.2}^7=-20\cdot -1.2800000000000005E-05=0.0002560000000000001$$

$$a_9=a_1·r^{n-1}=-20\cdot -{0.2}^{9-1}=-20\cdot -{0.2}^8=-20\cdot 2.5600000000000013E-06=-5.1200000000000025E-05$$

$$a_{10}=a_1·r^{n-1}=-20\cdot -{0.2}^{10-1}=-20\cdot -{0.2}^9=-20\cdot -5.120000000000002E-07=1.0240000000000005E-05$$