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The 3rd and 6th terms of an arithmetic progression are 94 and 85 respectively. Find the value of the 15th term.

 

From the 3rd term to the 6th term we have a 3 term skip, with a decrease of 9. Since this is an arithmetic progression, for every next term, we will subtract 3 from the pervious term since 9/3 =3.

15 - 6 = 9 more terms from the 6th term, so 9 x 3 = 27. Since the 6th term is 85, 85 - 27 = 58.

 

I hope this helped.