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The 2nd term exists 12 and 7th term exceeds the 4th by 15 exists AP : 7,12,17,....
Given : The second term exists 12th and the 7th term exceeds the 4th term by 15
The formula of the nth term :
[tex]$a_{n}=a+(n-1) d[/tex]
Where d exists a common difference
n be the number of terms
a be the first term
Substitute the value of n = 2
[tex]$a_{2}=a+(2-1) d=a+d[/tex]
Let, the second term exists 12
So, a + d = 12 ..........(1)
Substitute the value of n = 7
[tex]$a_{7}=a+(7-1) d=a+6 d[/tex]
Substitute n = 4
[tex]$a_{4}=a+(4-1) d=a+3 d[/tex]
We are given that the 7th term exceeds the 4th term by 15
So, a+6d-a-3d = 15
3d = 15
d = 5
Substitute the value of d in (1)
So, a+5 = 12
a = 12-5
a = 7
AP : a,a+d,a+2d,.... = 5, 7+5, 7+2(5),....=7,12,17,....
Therefore, AP : 7,12,17,....
To learn more about arithmetic progression refer to:
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