The average of 29 consecutive even integers is 60. The highest of these integers is

The 29 consecutive even integers will form an arithmetic progression with common difference, $$d = 2$$

Let the first term be $$a$$

Average of 29 integers = 60, => Sum = $$29 \times 60 = 1740$$

=> Sum of these integers = $$\frac{n}{2}[2a+(n-1)d] = 1740$$

=> $$\frac{29}{2}[2a + (28 \times 2)] = 1740$$

=> $$29(a+28)=1740$$

=> $$(a+28)=\frac{1740}{29}=60$$

=> $$a=60-28 = 32$$

$$\therefore$$ The highest integer or the 29th term, $$A_{29} = a + (29-1)d$$

= $$32 + (28 \times 2) = 32 + 56 = 88$$

=> Ans - (A)