Su'aalo Tusaale ah oo Ka Hadlaya Taxanaha Joomatari
Taxanaha joomatarigu waa fikrad aasaasi ah oo ku jirta xisaabta oo inta badan lagu barto dugsiga sare. Taxanahani waxay ka kooban yihiin tirooyin, mid walbana waa natiijada tiradii hore iyadoo la adeegsanayo joogto loo yaqaan "saamiga." Maqaalkani wuxuu dabooli doonaa dhowr tusaale oo dhibaatooyin ah iyo doodo ku saabsan taxanayaasha joomatari, iyadoo la rajeynayo inuu ka caawiyo akhristayaasha inay si fiican u fahmaan fikraddan.
Fahmidda Taxanaha Joomatari
Taxane joomatari waa taxane tirooyin ah oo lagu sameeyay ku dhufashada tirada koowaad (a) saamiga go'an (r). Guud ahaan, qaabka guud ee taxane joomatari waa:
\[ a, ar, ar^2, ar^3, \ldots, ar^{n-1} \]
Halkan:
- "a" waa ereyga ugu horreeya ee taxanaha.
- "r" waa saamiga hal erey iyo ereygii hore.
– "n" waa ereyga naad ee taxanaha.
Su'aalo iyo Doodo Tusaale ah
Aan ka wada hadalno qaar ka mid ah dhibaatooyinka tusaalaha ah si aan wax badan uga fahanno taxanaha joomatari.
Su'aal Tusaale 1aad
Su'aal:
Marka la eego taxane joomatari ah oo leh ereyga koowaad (a) waa 3 iyo saamiga (r) waa 2. Go'aami:
1. Muddada 5aad ee taxanaha.
2. Wadarta 6da erey ee ugu horreeya ee taxanaha.
Dood:
1. Ereyga 5aad (U5) waxaa lagu xisaabin karaa iyadoo la adeegsanayo qaacidada ereyga naad ee taxanaha joomatari, kuwaas oo kala ah:
\[ U_n = a \cdot r^{n-1} \]
Ku beddel a = 3, r = 2, iyo n = 5 qaacidada:
\[ U_5 = 3 \cdot 2^{5-1} \]
\[ U_5 = 3 \cdot 2^4 \]
\[ U_5 = 3 \cdot 16 \]
\[ U_5 = 48 \]
Markaa, ereyga 5aad ee taxanaha waa 48.
2. Wadarta 6da erey ee ugu horreeya (S6) ee taxanaha joomatari waxaa lagu xisaabin karaa iyadoo la adeegsanayo qaacidada wadarta ereyada n ee ugu horreeya, kuwaas oo kala ah:
\[ S_n = a \left( \frac{r^n – 1}{r – 1} \right) \]
Ku beddel a = 3, r = 2, iyo n = 6 qaacidada:
\[ S_6 = 3 \left( \frac{2^6 – 1}{2 – 1} \right) \]
\[ S_6 = 3 \left( \frac{64 – 1}{1} \right) \]
\[ S_6 = 3 \bidix(63 \midig) \]
\[ S_6 = 189 \]
Markaa, wadarta 6da erey ee ugu horreeya ee taxanaha waa 189.
Su'aal Tusaale 2aad
Su'aal:
Taxanaha joomatarigu wuxuu leeyahay erey 3aad oo ah 27 iyo erey 5aad oo ah 243. Go'aami qiimaha erey 1aad (a) iyo saamiga (r).
Dood:
Marka la eego U3 = 27 iyo U5 = 243. Iyadoo la adeegsanayo qaacidada ereyga naad ee taxanaha joomatari:
\[ U_n = a \cdot r^{n-1} \]
Loogu talagalay U3:
\[ U_3 = a \cdot r^2 \]
\[ 27 = a \cdot r^2 \] \[ (1) \]
Loogu talagalay U5:
\[ U_5 = a \cdot r^4 \]
\[ 243 = a \cdot r^4 \] \[ (2) \]
Isbarbardhigga isleegyada (1) iyo (2) si loo tirtiro a:
\[ \frac{U_5}{U_3} = \frac{a \cdot r^4}{a \cdot r^2} \]
\[ \frac{243}{27} = r^2 \]
\[ 9 = r^2 \]
\[ r = 3 \qoraal{ ama } r = -3 \]
Qiimaha r ku beddel isle'egta (1):
Haddii \( r = 3 \):
\[ 27 = a \cdot 3^2 \]
\[ 27 = a \cdot 9 \]
\[ a = 3 \]
Haddii \( r = -3 \):
\[ 27 = a \cdot (-3)^2 \]
\[ 27 = a \cdot 9 \]
\[ a = 3 \]
Markaa, ereyga koowaad (a) waa 3, saamiga (r)-na wuxuu noqon karaa 3 ama -3.
Su'aal Tusaale 3aad
Su'aal:
Soo hel wadarta aan dhammaadka lahayn ee taxanaha joomatari ee soo socda haddii ereyga koowaad (a) uu yahay 8 iyo saamiga (r) uu yahay 1/2.
Dood:
Wadarta aan dhammaadka lahayn ee taxanaha joomatari waxaa lagu xisaabin karaa qaacidada:
\[ S_{\infty} = \frac{a}{1 – r} \]
Ku beddel a = 8 iyo r = 1/2 qaacidada:
\[ S_{\infty} = \frac{8}{1 – \frac{1}{2}} \]
\[ S_{\infty} = \frac{8}{\frac{1}{2}} \]
\[ S_{\infty} = 8 \jeer 2 \]
\[ S_{\infty} = 16 \]
Markaa, wadarta aan dhammaadka lahayn ee taxanaha joomatarigu waa 16.
Su'aal Tusaale 4aad
Su'aal:
Taxanaha joomatarigu wuxuu leeyahay erey 2aad oo ah 12 iyo erey 4aad oo ah 108. Go'aami saamiga iyo erey 1aad ee taxanaha.
Dood:
La bixiyay \( U_2 = 12 \) iyo \( U_4 = 108 \). Iyadoo la adeegsanayo qaacidada ereyga naad ee taxanaha joomatari:
Wixii \( U_2 \):
\[ U_2 = a \cdot r \]
\[ 12 = a \cdot r \] \[ (1) \]
Wixii \( U_4 \):
\[ U_4 = a \cdot r^3 \]
\[ 108 = a \cdot r^3 \] \[ (2) \]
Isbarbardhigga isleegyada (1) iyo (2) si loo tirtiro a:
\[ \frac{U_4}{U_2} = \frac{a \cdot r^3}{a \cdot r} \]
\[ \frac{108}{12} = r^2 \]
\[ 9 = r^2 \]
\[ r = 3 \qoraal{ ama } r = -3 \]
Qiimaha r ku beddel isle'egta (1):
Haddii \( r = 3 \):
\[ 12 = a \cdot 3 \]
\[ a = 4 \]
Haddii \( r = -3 \):
\[ 12 = a \cdot (-3) \]
\[ a = -4 \]
Markaa, ereyga koowaad (a) wuxuu noqon karaa 4 ama -4, saamiga (r) wuxuu noqon karaa 3 ama -3.
Gabagabo
Taxanaha joomatarigu waa fikrad xisaabeed oo muhiim ah oo inta badan loo isticmaalo dhinacyo kala duwan. Markaan fahamno aasaaska iyo ku celcelinta xirfadaha xallinta mashaakilaadka, waxaan rajeyneynaa inaan noqono kuwo aad ugu xeel dheer fahamka iyo hirgelinta fikradda. Maqaalkani wuxuu ka kooban yahay dhowr tusaale oo dhibaatooyin iyo doodo ah si looga caawiyo akhristayaasha inay si qoto dheer u bartaan oo ay u fahmaan taxanaha joomatari. Waxaan rajeyneynaa inay tani waxtar leedahay!
