For the AP: x−7,x−2,x+3,.... the 15th term is ____.
x+63
Given AP is: x−7,x−2,x+3,....
∴ First term (a)=x−7
Common difference: (d)=(x−2)−(x−7)=5
The nth term of an AP is given by, an=a+(n−1)d
Then, the 15th term of the above series is given by,
a15=(x−7)+(15−1)5
⇒a15=x−7+70 =x+63