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

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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,.....

Posted by

Gautam harsolia

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