Q : 16      Determine the AP whose third term is \small 16 and the \small 7th term exceeds the \small 5th term by \small 12.
 

Answers (1)

It is given that
3rd term of AP is \small 16 and the \small 7th term exceeds the \small 5th term by \small 12
i.e.
a_3=a+2d = 16 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)
And 
a_7=a_5+12
a+6d=a+4d+12
2d = 12
d = 6
Put the value of d in equation (i) we will get
a = 4
Now, AP with first term = 4 and common difference = 6 is 
4,10,16,22,.....

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