the sum of 1st n terms of an AP is given by the formula Sn=3n^2+n, then its 3rd term is
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S
n
=3n
2
+2n
Taking n=1, we get
S
1
=3(1)
2
+2(1)
⇒S
1
=3+2
⇒S
1
=5
⇒a
1
=5
Taking n=2, we get
S
2
=3(2)
2
+2(2)
⇒S
2
=12+4
⇒S
2
=16
∴a
2
=S
2
−S
1
=16−5=11
Taking n=3, we get
S
3
=3(3)
2
+2(3)
⇒S
3
=27+6
⇒S
3
=33
∴a
3
=S
3
−S
2
=33−16=17
So, a=5,
d=a
2
−a
1
=11−5=6
Now, we have to find the 15
th
term
a
n
=a+(n−1)d
a
15
=5+(15−1)6
a
15
=5+14×6
a
15
=5+84
a
15
=89
Hence, the 15
th
term is 89 and AP is 5,11,17,23,...
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