In a certain sequence, the term \(a_n\) is defined as the value of x that satisfies the equation \(2 = \frac{x}{2} -a_{n- 1}\). If \(a_6 =156\), what is the value of \(a_2\)?

(A) 1
(B) 6
(C) 16
(D) 26
(E) 106

\(2 = \frac{x}{2} -a_{n- 1}\)

\( \frac{x}{2} =2+a_{n- 1}\)

\(x=a_n=2+a_{n-1}\), as x is nothing but \(a_n\).

Two ways

I. Find the lower terms from the given value
\(a_n=4+2*a_{n-1}\)
\(a_6=4+2*a_{5}\)
\(156=4+2*a_5……..a_5=76\)

\(a_5=4+2*a_{4}\)
\(76=4+2*a_4……..a_4=36\)

\(a_4=4+2*a_{3}\)
\(36=4+2*a_3……..a_3=16\)

\(a_3=4+2*a_{2}\)
\(16=4+2*a_2……..a_2=6\)


II. Use options.

A. 1…. \(a_3=4+2*a_{2}=4+2=6……… a_4=4+2*a_{3}=4+12=16……….. a_5=4+2*a_{4}\)
\(=4+32=36……….. a_6=4+2*a_{5}=4+72=76\neq 156\)

B. 6…. \(a_3=4+2*a_{2}=4+12=16……… a_4=4+2*a_{3}=4+32=36……….. a_5=4+2*a_{4}\)
\(=4+72=76……….. a_6=4+2*a_{5}=4+76=156\)…..Correct

B