Let X be the set consisting of the first 2018 term of the arithmetic progression 1, 6, 11, .………… , and Y ‘be the set consisting of the first 2018 terms of the arithmetic progression 9, 16, 23, ……….. Then, the number of elements in the set X∪Y is

# Let be the set consisting of the first 2018 term of the arithmetic progression 1, 6, 11, .………… , and $Y$ 'be the set consisting of the first 2018 terms of the arithmetic progression 9, 16, 23, ……….. Then, the number of elements in the set $X\cup Y$ is

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### Solution:

Here $X=\left\{1,6,11,\dots ,10086\right\}$

and $Y=\left\{9,16,23,\dots ,14128\right\}$

The intersection of $X$ andis an A.P. with 16 as first term and 35 as common difference.

The series becomes 16, 51, 86, ..

Now,  term =$16+\left(k-1\right)35\le 10086$

i.e. $k\le \frac{10105}{35}\therefore k\le 288$ (as is to be an integer)

Hence,$\begin{array}{r}n\left(X\cup Y\right)=n\left(X\right)+n\left(Y-n\left(X\cap Y\right)\\ =2018+2018-288=3748\end{array}$

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