Determine the A.P. whose third term is 16 and 7th term exceeds the
5th term by 12.please show the method it's urgent
Answers
Answered by
1
Answer:
Yes It Is An A.P
4,10,16,22,28,34,40.
a=4,d=6
Step-by-step explanation:
t7th = t5th + 12
a+6d=a+4d+12
6d-4d=12
2d=12
d=6.
t3th = 16
a+2(d)=16
a+2(6)=16
a=16-12
a=4.
Answered by
1
Answer:
The answer is 4, 10, 16, 22,......
Step-by-step explanation:
Since the third term is given to us as 16,
=> a+2d = 16 (An = a+(n-1)d) ....(1)
Also, the difference between the 7th term and the 5th term is 12
So,
a + 6d - (a + 4d) = 12
a + 6d - a - 4d = 12
2d = 12
d = 6
Now putting d = 6 in (1)
a + 12 = 16
a = 4
So the AP becomes 4, 10(4+6), 16(4 + 6×2) , 22 and so on
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