1. The average of 19 consecutive even integers is 50. The highest of these integers is
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By: anil on 05 May 2019 02.15 am
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 = $$19 imes 50 = 950$$ => Sum of these integers = $$frac{n}{2}[2a+(n-1)d] = 950$$ => $$frac{19}{2}[2a + (18 imes 2)] = 950$$ => $$19(a+18)=950$$ => $$(a+18)=frac{950}{19}=50$$ => $$a=50-18 = 32$$ $$ herefore$$ The highest integer or the 19th term, $$A_{19} = a + (19-1)d$$ = $$32 + (18 imes 2) = 32 + 36 = 68$$ => Ans - (B)
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