In a certain sequence, the term an is defined by the formula an = an –

\(a_n\) is defined by the formula \(a_n = a_{n – 1} + 5\) for each integer n ≥ 2. If \(a_1 = 1\),

\(A_1=1 ------>A_1=1\)
\(A_2=A_1+5=6->A_2=1+5\)
\(A_3=A_2+5=11->A_3=1+5+5\)\(A_4=A_3+5=16->A_4\\
=1+5+5+5\)

The sequence on the right just helps to show that this is an arithmetic series with a common difference of 5 and a start term of 1, that is
\(A_n=1+(n-1)5\)

\(A_2\) is 1 + one 5
\(A_3\) is 1 + two 5s
\(A_4\) is 1 + three 5s
\(A_5\) will be 1 + four 5s
\(A_{6}\) = 1 + five 5s

Each term starts with \(A_1=1\)
Each term has a common difference of 5 (increases by 5) -- BUT
Each Term\(_{n}\) has \((n-1)\) 5s.
Each term's number of 5s is one fewer than itself.
\(A_{75}\) will have \((n-1) = 74\) fives

\(A_n=1+(n-1)5\): sequence rule from above

\(A_{75}=(1+(75-1)*5)=\)
\(A_{75}=(1+(74*5))\)
\(A_{75}=(1+370)=371\)

Sum of the sequence: (average)*(# of terms)

Average in an evenly spaced sequence is \(\frac{(First+ Last)}{2}\)

# of terms, given, is 75

Sum: \(\frac{(First+ Last)}{2}*75\)

Sum = \((\frac{1+371}{2}*75)=(186*75)\)
Double and halve
Sum = \((93*150)=\)
Split the 93
Sum of sequence =\((90*150+3*150)=(13,500+450)= 13,950\)

Answer D

For an excellent post on sequences that is easy to follow, see this topic thread on GMAT Club, and scroll down to the post by benjiboo

See also the Ultimate GMAT Quantitative Megathread , 12. Sequences