Question

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Answer 1

n−2,4n−1,5n+2 are in A.P.

First term, a1  =n−2

Second term, a 2  =4n−1

Third term, a3   =5n+2

Common difference (difference of two consecutive terms),  

d= a 2 − a1 = a3 − a2

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,  

a1 = n−2 = 1−2 = −1

a2 = 4n−1 = 4−1=3

and d=a2  −a1  = 3−(−1)=4

nth term of an AP is ,

an =a+(n−1)d

where,

a=first term

d=common difference

Therefore,

a4 =a+3d=−1+12=11

a5 =a+4d=−1+16=15

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