Question

Find the first five term of following given sequences: am = (2m2

) - 3m + 1

Answer 1

Answer:

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Answer 2

$Given \: \red{ a_{m} = 2m^{2} - 3m + 1 }$

$i) First \:term (a_{1}) = 2(1)^{2} - 3(1) + 1 \\= 2 - 3 + 1 \\= 3 - 3 \\= 0$

$ii) Second \:term (a_{2}) = 2(2)^{2} - 3(2) + 1 \\= 8 - 6 + 1 \\= 2+1 \\= 3$

$iii) Third \:term (a_{3}) = 2(3)^{2} - 3(3) + 1 \\= 18 - 9 + 1 \\= 9+1 \\= 10$

$iv) Fourth \:term (a_{4}) = 2(4)^{2} - 3(4) + 1 \\= 32 - 12 + 1 \\= 20+1 \\= 21$

$v) Fifth \:term (a_{5}) = 2(5)^{2} - 3(5) + 1 \\= 50 - 15 + 1 \\= 35+1 \\= 36$

Therefore.,

$\green { 0, 3,10,21 \:and \: 36 \:are \: first}\\ \green {five \:terms \: of \: given \: sequence }$

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