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શ્રેણી અને શ્રેઢી

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ગણિત ધોરણ 11 સેમિસ્ટર 2

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ગણિત

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નીચેની શ્રેણીના પ્રથમ પાંચ પદો લખો.
f(n) = 3n + 1


અહીં, f(n) = 3n + 1

તેથી f(1) = 3 (1) + 1 = 4
   
     f(2) = 3 (2) + 1 = 7

 
 
     f(3) = 3 (3) + 1 = 10

 
    f(4) = 3 (4) + 1 = 13


    f(5) = 3 (5) + 1 = 16

પ્રથમ પાંચ પદ : 4 , 7 , 10, 13, 16


ફિનોબાકી શ્રેણી straight a subscript 1 space equals space straight a subscript 2 space equals space 1 space અન ે space straight a subscript straight n space equals space straight a subscript straight n minus 1 end subscript space plus space straight a subscript straight n space minus space 2 comma space end subscript straight n space greater than space 2 હોય, તો straight a subscript 3 comma space straight a subscript 4 comma space straight a subscript 5 comma space straight a subscript 6 શોધો.

 


અહીં, straight a subscript 1 space equals space straight a subscript 2 space equals space 1 આપેલું છે.

અને straight a subscript straight n space equals space straight a subscript straight n minus 1 end subscript space plus space straight a subscript straight n space minus space 2 end subscript space comma space straight n space greater than space 2

આથી

straight a subscript 3 space equals space straight a subscript 2 space plus space straight a subscript 1 space equals space 1 space plus space 1 space equals space 2

straight a subscript 4 space equals space straight a subscript 3 space plus space straight a subscript 2 space equals space 2 space plus space 1 space equals space 3

straight a subscript 5 space equals space straight a subscript 4 space plus space straight a subscript 3 space equals space 3 space plus space 2 space equals space 5

straight a subscript 6 space equals space straight a subscript 5 space plus space straight a subscript 4 space equals space 5 space plus space 3 space equals space 8


નીચેની શ્રેણીના પ્રથમ પાંચ પદો લખો.
f left parenthesis straight n right parenthesis space equals space fraction numerator straight n space minus space open parentheses negative 1 close parentheses to the power of straight n over denominator 2 end fraction

 


અહીં, f left parenthesis n right parenthesis space equals space fraction numerator n space minus space open parentheses negative 1 close parentheses to the power of n over denominator 2 end fraction

તેથી 

f open parentheses italic 1 close parentheses italic space italic equals italic space fraction numerator italic 1 italic space italic minus italic space open parentheses italic minus italic 1 close parentheses to the power of italic 1 over denominator italic 2 end fraction italic space italic equals italic space fraction numerator italic 1 italic space italic plus italic space italic 1 over denominator italic 2 end fraction italic space italic equals italic space italic 1

f open parentheses italic 2 close parentheses italic space italic equals italic space fraction numerator italic 2 italic space italic minus italic space open parentheses italic minus italic 1 close parentheses to the power of italic 2 over denominator italic 2 end fraction italic space italic equals italic space fraction numerator italic 2 italic space italic minus italic space italic 1 over denominator italic 2 end fraction italic space italic equals italic space italic 1 over italic 2

f open parentheses italic 3 close parentheses italic space italic equals italic space fraction numerator italic 3 italic space italic minus italic space open parentheses italic minus italic 1 close parentheses to the power of italic 3 over denominator italic 2 end fraction italic space italic equals italic space fraction numerator italic 3 italic space italic plus italic space italic 1 over denominator italic 2 end fraction italic space italic equals italic space italic 2

f open parentheses italic 4 close parentheses italic space italic equals italic space fraction numerator italic 4 italic space italic minus italic space open parentheses italic minus italic 1 close parentheses to the power of italic 4 over denominator italic 2 end fraction italic space italic equals italic space fraction numerator italic 4 italic space italic minus italic space italic 1 over denominator italic 2 end fraction italic space italic equals italic space italic 3 over italic 2

f open parentheses italic 5 close parentheses italic space italic equals italic space fraction numerator italic 5 italic space italic minus italic space open parentheses italic minus italic 1 close parentheses to the power of italic 5 over denominator italic 2 end fraction italic space italic equals fraction numerator italic 5 italic space italic plus italic space italic 1 over denominator italic 2 end fraction italic space italic equals italic space italic 3

પ્રથમ પાંચ પદ : bold 1 bold comma bold space bold 1 over bold 2 bold comma bold space bold 2 bold comma bold space bold 3 over bold 2 bold comma bold space bold 3


નીચેની શ્રેણી માટે straight a subscript 2 comma space straight a subscript 3 comma space straight a subscript 4 શોધો.
straight a subscript 1 space equals space minus 3 space અન ે space straight a subscript straight n space equals space 2 straight a subscript straight n space minus space 1 end subscript space plus space 1 comma space for all straight n space greater than space 1

 


અહીં, straight a subscript 1 space equals space minus 3 આપેલું છે.

અને straight a subscript straight n space equals space 2 straight a subscript straight n space minus space 1 end subscript space plus space 1 comma space for all straight n space greater than space 1.

આથી

straight a subscript 2 space equals space 2 straight a subscript 1 space plus space 1 space equals space 2 space open parentheses negative 3 close parentheses space plus space 1 space equals space minus 5

straight a subscript 3 space equals space 2 straight a subscript 2 space plus space 1 space equals space 2 space open parentheses negative 5 close parentheses space plus space 1 space equals space minus 9

straight a subscript 4 space equals space 2 straight a subscript 3 space plus space 1 space equals space 2 space open parentheses negative 9 close parentheses space plus space 1 space equals space minus 17


નીચેની શ્રેણીના પ્રથમ પાંચ પદો લખો.
f(n) = nમી અવિભાજ્ય સંખ્યા

 


અહીં, f(n) = nમી અવિભાજ્ય સંખ્યા

તેથી 

f(1) = પહેલી અવિભાજ્ય સંખ્યા = 2

f(2) = બીજી અવિભાજ્ય સંખ્યા = 3

f(3) = ત્રીજી અવિભાજ્ય સંખ્યા = 5

f(4) = ચોથી અવિભાજ્ય સંખ્યા = 7

f(5) = પાંચમી અવિભાજ્ય સંખ્યા = 11

પ્રથમ પાંચ પદ : 2. 3, 5, 7, 11