In a arithmetic progression a1=3 and a6= 43 find the first 5 numbers

a=3

a6=43

a+6d=431

substituting a=3 in eq 1

a+6d=43

3+6d=43

6d=43-3

d=40/6

d=20/3

so first 5 no's are a, a+d, a2+d,a3+d, a4+d

a=3

a+d=3+20/d

i think u know after this

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3,11,19,27,35

Step-by-step explanation:

an=a1+(n-1)d

a1=3 (given)

a6=43

d=?

a6= a1+(6-1)d

43= 3 + 5d

d= 8

first number a1=3

a2= a1+d

a2= 3+8=11

a3= a1+2d

= 3+2(8)=19

a4= a1+3d

= 3+3(8)=27

a5=a1+4d

=3+4(8)=35