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secondary 2 | Maths
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Date Posted: 1 month ago
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t7
= t6 + t5 = (t5 + t4) + t5
= 2(t5) +t4 = 2(t4 + t3) +t4
= 3(t4) + 2(t3) = 3(t3 + t2) + 2(t3)
= 5(t3) + 3(t2) = 5(t2 + t1) + 3(t2)
= 8(t2) + 5(t1) = 184
-> (t1) = 1/5 x [184 - 8(t2)]
Since both t1 and t2 are positive integers
-> [184 - 8(t2)] must be a positive multiple of 5
-> t2 = 3, 8, 13, 18 and correspondingly
-> t1 = 32, 24, 16, 8
Since t1 < t2
-> t1 = 8 and t2 = 18
t12
= t11 + t10
= 2(t10) +t9
= 3(t9) + 2(t8)
= 5(t8) + 3(t7)
= 8(t7) + 5(t6)
= 8(184) + 5(t5) + 5(t4)
= 1,472 + 10(t4) + 5(t3)
= 1,472 + 15(t3) + 10(t2)
= 1,472 + 25(t2) + 10(t1)
= 1,472 + 25(18) + 15(8)
= 2,042
= t6 + t5 = (t5 + t4) + t5
= 2(t5) +t4 = 2(t4 + t3) +t4
= 3(t4) + 2(t3) = 3(t3 + t2) + 2(t3)
= 5(t3) + 3(t2) = 5(t2 + t1) + 3(t2)
= 8(t2) + 5(t1) = 184
-> (t1) = 1/5 x [184 - 8(t2)]
Since both t1 and t2 are positive integers
-> [184 - 8(t2)] must be a positive multiple of 5
-> t2 = 3, 8, 13, 18 and correspondingly
-> t1 = 32, 24, 16, 8
Since t1 < t2
-> t1 = 8 and t2 = 18
t12
= t11 + t10
= 2(t10) +t9
= 3(t9) + 2(t8)
= 5(t8) + 3(t7)
= 8(t7) + 5(t6)
= 8(184) + 5(t5) + 5(t4)
= 1,472 + 10(t4) + 5(t3)
= 1,472 + 15(t3) + 10(t2)
= 1,472 + 25(t2) + 10(t1)
= 1,472 + 25(18) + 15(8)
= 2,042
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