WJEC Unit 3 2018 June — Question 8 5 marks

Exam BoardWJEC
ModuleUnit 3 (Unit 3)
Year2018
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeFind term or common difference
DifficultyModerate -0.3 This is a straightforward application of arithmetic progression formulas. Given n=7, the last term a+6d=71, and sum S_7=329, students can use S_n=n/2(first+last) to find the first term immediately (329=7/2(a+71) gives a=23), then solve for d. The question requires only direct formula application with no conceptual challenges or problem-solving insight.
Spec1.04h Arithmetic sequences: nth term and sum formulae

Find seven numbers which are in arithmetic progression such that the last term is 71 and the sum of all of the numbers is 329. [5]

Find seven numbers which are in arithmetic progression such that the last term is 71 and the sum of all of the numbers is 329. [5]

\hfill \mbox{\textit{WJEC Unit 3 2018 Q8 [5]}}