a3=18 a7=30 find s17

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Solution

a3=18
a7=30
s17=?
a3=a+2d=18 (1)
a7=a+6d=30 (2)
on substracting (1) from (2), we get ,
-4d=-12
d=3
putting the value of d in equation (1) we get,
a+2(3)=18
a+6=18
a=18-6
a=12
now a=12 and d=3
now,
s17=n/2[2a+(n-1)d]
s17= 17/2[2(12)+(17-1)(3)]
s17=17/2[24+48]
s17=17/2(72)
s17=612
therefore, the sum of the 17th term of an a.p is 612

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