If 2nd term of an AP is 13 and 5th is 25.What is 9th term?
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Thus the answer is 33.
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a2=13 and a5=25
We know that
an=a+(n−1)d
∴a2=a+(2−1)d
⇒13=a+(2−1)d
⇒a+d=13 ...(i)
and a5=a+(5−1)d
⇒25=a+4d
⇒a+4d=25
(ii)Now, subtracting
(i) from
(ii), we get
3d=12
⇒d=4
Now a+d=13 [from (i)]
⇒a=9
Hence, a7=a+6d=9+6(4)
⇒a7=33
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