If (a_n=4n^2-3n+8), what are the first three terms?
Answer and explanation
Correct answer: (9,18,35)
Substitute n=1, 2, and 3 successively. This gives a_1=4(1)^2-3(1)+8=9, a_2=4(2)^2-3(2)+8=18, and a_3=4(3)^2-3(3)+8=35. Hence, the correct sequence is (9,18,35). Option (9,20,39) has the correct first term, but its calculations for n=2 and n=3 are incorrect. Exam tip: substitute each value of n separately, carrying out squaring and multiplication first.
Frequently asked questions
What is the correct answer to this question?
(9,18,35)
Why is this the correct answer?
Substitute n=1, 2, and 3 successively. This gives a_1=4(1)^2-3(1)+8=9, a_2=4(2)^2-3(2)+8=18, and a_3=4(3)^2-3(3)+8=35. Hence, the correct sequence is (9,18,35). Option (9,20,39) has the correct first term, but its calculations for n=2 and n=3 are incorrect. Exam tip: substitute each value of n separately, carrying out squaring and multiplication first.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: Explicit or general rule.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
