Let $x_{1}, x_{2}, . . . . .$ be positive integers in A.P., such that $x_{1} + x_{2} + x_{3} = 12$ and $x_{4} + x_{6} = 14$ . Then $x_{5}$ is-

A prime number

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Solution

The correct options are
A A prime number
D 7
Let $d$ be the common difference of an A.P. & nth term is given by ( first term + (n-1)d )
so, $x_{2} = x_{1} + d$
$x_{3} = x_{1} + 2 d$
$x_{4} = x_{1} + 3 d$
$x_{5} = x_{1} + 4 d$
$x_{6} = x_{1} + 5 d$
Now, $x_{1} + x_{2} + x_{3} = 12$
$\Rightarrow x_{1} + x_{1} + d + x_{1} + 2 d = 12$
$\Rightarrow x_{1} + d = 4$ $- - - - - (1)$
Also, $\Rightarrow x_{4} + x_{6} = 14$
$\Rightarrow x_{1} + 3 d + x_{1} + 5 d = 14$
$\Rightarrow x_{1} + 4 d = 7$ $- - - - - (2)$
Now, Subtract equation $(2)$ by $(1)$ ; we get $d = 1$ .
Substitute $d$ in equation $(1)$ ; we get $x_{1} = 3$ .
Therefore, $x_{5} = 7$ which is a prime number.

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Let $x_{1}, x_{2}, . . . . .$ be positive integers in A.P., such that $x_{1} + x_{2} + x_{3} = 12$ and $x_{4} + x_{6} = 14$ . Then $x_{5}$ is-