Find two consecutive positive integers such that the square of the larger integer is nineteen more than nine times the smaller i
1 answer:
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Answer:
see explanation
Step-by-step explanation:
These are the terms of an arithmetic sequence with n th term
= a + (n - 1)d
where a is the first term and d the common difference
d = 25 - 20 = 30 - 25 = 5 and a = 20, hence
= 20 + 5(n - 1) = 20 + 5n - 5 = 5n + 15 ← n th term formula