Question 43

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

Solution

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)


Create a FREE account and get:

  • Free SSC Study Material - 18000 Questions
  • 230+ SSC previous papers with solutions PDF
  • 100+ SSC Online Tests for Free

cracku

Boost your Prep!

Download App