In an ap the forth term exceeds four times the 12th term by one and third term exceeds twice the tenth term by 5 find ap
Answers
let the first term of ap be "a" and common difference be"d".
following the first direction secondly
a4+1=4(a12) a3+5=2(a10)
a+3d+1=4(a+11d) a+2d+5=2(a+9d)
3a +41d=1........(1) a+16d=5.................(2)
substracting (2) from (1) we get
d=2
a= -27
therefore Ap is -27,-25,-23,-21,......
probably this is the answer
Answer:
According to the question
a4=4a12+1a4=4a12+1
a1+3d=4(a1+11d)+1a1+3d=4(a1+11d)+1
a1+3d=4a1+44d+1a1+3d=4a1+44d+1
−3a1=41d+1−3a1=41d+1
−a1=41d+13−a1=41d+13 ............(1)
a3=2a10+5a3=2a10+5
a1+2d=2(a1+9d)+5a1+2d=2(a1+9d)+5
a1+2d=2a1+18d+5a1+2d=2a1+18d+5
−a1=16d+5−a1=16d+5
From eq (1)
41d+13=16d+541d+13=16d+5
41d+1=48d+1541d+1=48d+15
−7d=14−7d=14
d=−2d=−2
a1=41d+1−3=41×−2+1−3=21a1=41d+1−3=41×−2+1−3=21
∴∴ series in AP =27,25,23,21
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