If the no. N-2 ,4n-1 and 5n+2 are in AP ,then the value of n is
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n−2,4n−1,5n+2 are in A.P.
First term, a
1
=n−2
Second term, a
2
=4n−1
Third term, a
3
=5n+2
Common difference (difference of two consecutive terms),
d=a
2
−a
1
=a
3
−a
2
4n−1−(n−2)=5n+2−(4n−1)
4n−1−n+2=5n+2−4n+1
3n+1=n+3
2n=2
n=1
So,
a
1
=n−2=1−2=−1
a
2
=4n−1=4−1=3
and d=a
2
−a
1
=3−(−1)=4
nth term of an AP is ,
a
n
=a+(n−1)d
where,
a=first term
d=common difference
Therefore,
a
4
=a+3d=−1+12=11
a
5
=a+4d=−1+16=15
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