JEE MainMathematicsEasySeries and progressions

Let 6 Arithmetic means $A_{1},A_{2},A_{3},A_{4},A_{5},A_{6}$ are
inserted between two consecutive natural number a and (a > b). If
$A_{1}^{2} - A_{2}^{2} + A_{3}^{2} - A_{4}^{2} + A_{5}^{2} - A_{6}^{2}$
is equal to prime number then is equal to

Show answer and solution

b_{1}\ A_{1},\ \ A_{2},\ \ A_{3}\ \cdots\cdots\cdots,a

A_{1} = b + d

A_{2} = b + 2d

\begin{matrix} \because & a = b + 1 \end{matrix}

\Rightarrow a - b = 1

\Rightarrow d = \frac{a - b}{7} = \frac{1}{7}

\Rightarrow A_{1} = b + \frac{1}{7}

A_{2} = b + \frac{2}{7}

\Rightarrow \ A_{1}^{2} - A_{2}^{2} + A_{3}^{2} - A_{4}^{2} + A_{5}^{2} - A_{6}^{2}

= \left( A_{1} + A_{2} \right)\left( A_{1} - A_{2} \right) + \left( A_{3} - A_{4} \right)\left( A_{3} + A_{4} \right) + \left( A_{5} - A_{6} \right)\left( A_{5} + A_{6} \right)

= \left( 2b + \frac{3}{7} \right)\frac{1}{7} + \left( 2b + \frac{7}{7} \right)\left( \frac{1}{7} \right) + \left( 2b + \frac{11}{7} \right)\frac{1}{7}

= \frac{1}{7}\left( 6b + \frac{3 + 7 + 11}{7} \right)

= \frac{1}{7}(6b + 3) = \frac{3}{7}(2b + 1) \Rightarrow

Prime number

\Rightarrow (2b + 1)

should be multiple of 7

\Rightarrow 2b + 1 = 7

\Rightarrow b = 3

Source: Grass-G / M-06 Mathematics Paper 14 October.docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.