In an A.P the fourth term exceed four times the 12th term by one and the third term exceeds twice than 10 th term by five find the ap
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Answer:
According to the question
a
4
=4a
12
+1
a
1
+3d=4(a
1
+11d)+1
a
1
+3d=4a
1
+44d+1
−3a
1
=41d+1
−a
1
=
3
41d+1
............(1)
a
3
=2a
10
+5
a
1
+2d=2(a
1
+9d)+5
a
1
+2d=2a
1
+18d+5
−a
1
=16d+5
From eq (1)
3
41d+1
=16d+5
41d+1=48d+15
−7d=14
d=−2
a
1
=
−3
41d+1
=
−3
41×−2+1
=21
∴ series in AP =27,25,23,21
Thank you❤
#kanishka.... ❤
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