You are given a sequence of numbers: -2, 12, 32, 58, 90, ... Work out the 7th term in this sequence.

This is a question from Higher Tier and involves 1) realising that 2nd difference is a constant: 6. 2) knowing that constant 2nd difference implies the sequence is quadratic. 3) remembering or proving that constant of 2nd difference - "a" gives us the coefficient of n^2 in the sequence as (a/2)n^2. In our case the answer is (3)n^2. 4) Working out the whole sequence by calculating 3n^2 for n = 1, 2, ... 5 and subtracting the original sequence from the latter. 5) obtaining the linear sequence from the difference to be -5, 0, 5, 10, ... and calculating its (n)th term to be (5n - 10) for n > 0. 6) Merge the two (n)th term sequences into [3n^2 + 5n - 10]. 7) substitute 7 into the whole sequence to obtain 3*(7^2) + 5(7) - 10 = 172. 

Answered by Maxym V. Maths tutor

4371 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Solve the equation (x+2)/(x-3)=(x-6)/(x+1) for x.


How do you solve the equation 3(6x + 2) = 10 + 4x?


The point P has coordinates (3, 4) The point Q has coordinates (a, b) A line perpendicular to PQ is given by the equation 3x + 2y = 7 Find an expression for b in terms of a.


a=7 and b=2, Work out the value of (a/b)-a^b


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy
Cookie Preferences