The seventh term of an arithmetic progression is four times it's second and twelth term is two more than three times of its fourth term. Find the progression.
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Step-by-step explanation:
a7=4(a2)
a12=2+3(a4)
a+6d=a+d(4)
a+6d=4a+4d
6d-4d=4a-a
2d=3a_____1
a+11d=3(a4)+2
a+11d=3(a+3d)+2
a+11d=3a+9d+2
11d-9d=3a-a+2
2d=2a+2_______2
substitute value 1in 2
3a=2a+2
3a-2a=2
a=2_____3
3a=2d
3(2)=2d
6=2d
d=6/2
d=3
therefore 2,5,8,11
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