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Yogita Ingle 6 years, 3 months ago
An = 5n-2
First term,a = 5×1-2 = 3
second term,a 2 = 5×2-2 = 8
d = a2-a1 = 8-3 = 5
12th term = 5×12 - 2= 60-2 = 58
15th term = 5×15 - 2= 75 - 2 = 73
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Yogita Ingle 6 years, 3 months ago
an = 24 - 3n
We need the second term and therefore :
n = 2
We substitute this value of n in the formula to get :
a2 = 24 - 3 × 2
= 24 - 6 = 18
The second term is thus 18.
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Yogita Ingle 6 years, 3 months ago
In ∆AOB, draw a perpendicular line from O which intersect AB at M.
In ∆AOM, angle AMO = 90
angle OAM = 30
cos 30 = AM/AO
√3/2 = AM/21
AM = 21×√3/2
AB = 2(AM)
=2(21×√3/2)
=21√3
{tex}OM^2 = AO^2-AM^2{/tex}
{tex}=21^2-(21√3/2)^2{/tex}
=441-330.51
=110.48
OM =√110.48
OM =10.51
OM = 10.51cm
Area of ∆AOM = 1/2 AB × OM
=1/2 ×21√3 ×10.51
=191.14cm^2
Area of sector AOBY = 120πr^2/360
=120×21×21×22/2520
=462cm^2
Area of segment AYB = Area of sector OAYB -Area of∆OAB
=462-191.14
=270.86
Area of segment AYB is 270.86cm2.
Posted by Bhoomi Dubey 6 years, 3 months ago
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Yogita Ingle 6 years, 3 months ago
HCF of 210 and 55
255= 55×3+45...(1)
55=45×1+10...(2)
45=10×4+5...(3)
10=5×2+0. remainder =0
HCF of 210 and 55 =5
NOW,from (3) we get,
45=10×4+5
45-10×4=5
45-(55-45)×4 =5 ....(from 2)
45-55×4+45×4=5
(210-55×3)5-55×4=5 ...(from 1)
210×5-55×15-55×4=5
5=210×5-55×19=210A+55B
therefore,
a=5 and b= -19
Posted by Chirag Salwan 6 years, 3 months ago
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Dips ? Tyagi 6 years, 3 months ago
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