Standard +0.3 This is a straightforward algebraic manipulation question using standard arithmetic series formulas. Students need to apply S_n = n/2[2a + (n-1)d] for n=36 and n=6, set up the equation S_36 = (S_6)^2, then expand and simplify. While it requires careful algebra across multiple steps, it involves only direct application of a memorized formula with no problem-solving insight or novel approach needed, making it slightly easier than average.
3.
An arithmetic sequence has first term \(a\) and common difference \(d\).
The sum of the first 36 terms of the sequence is equal to the square of the sum of the first 6 terms.
Show that \(4 a + 70 d = 4 a ^ { 2 } + 20 a d + 25 d ^ { 2 }\)
[0pt]
[4 marks]
3.
An arithmetic sequence has first term $a$ and common difference $d$.\\
The sum of the first 36 terms of the sequence is equal to the square of the sum of the first 6 terms.
Show that $4 a + 70 d = 4 a ^ { 2 } + 20 a d + 25 d ^ { 2 }$\\[0pt]
[4 marks]\\
\hfill \mbox{\textit{SPS SPS SM 2021 Q3 [4]}}