The average of 19 consecutive even integers is 50. The highest of these integers is

The 19 consecutive even integers will form an arithmetic progression with common difference, $$d = 2$$

Let the first term be $$a$$

Average of 19 integers = 50, => Sum = $$19 \times 50 = 950$$

=> Sum of these integers = $$\frac{n}{2}[2a+(n-1)d] = 950$$

=> $$\frac{19}{2}[2a + (18 \times 2)] = 950$$

=> $$19(a+18)=950$$

=> $$(a+18)=\frac{950}{19}=50$$

=> $$a=50-18 = 32$$

$$\therefore$$ The highest integer or the 19th term, $$A_{19} = a + (19-1)d$$

= $$32 + (18 \times 2) = 32 + 36 = 68$$

=> Ans - (B)