Sequences

(n(n+1)^2) does not give the full sequence, so care is needed. None of the given options matches all terms.

This is (3^2,4^2,5^2,\ldots), so the (20)th term is (22^2=484). Exam tip: track how the index shifts.

This is (2^3,3^3,4^3,\ldots), so the (10)th term is (11^3=1331). Exam tip: handle shifted cube sequences carefully.

The rule is: next term = twice the previous term − 1. Therefore, the term after 33 is \(2\times 33-1=66-1=65\). Getting 63 would mean subtracting 3 from 66, but the rule requires subtracting only 1. Exam tip: in a recursive sequence, apply the stated rule directly to the last given term.

The rule is: next term = twice the previous term + 1. Therefore, the term after 47 is \(2\times47+1=94+1=95\). Option 94 is only twice 47; it misses the required addition of 1. Exam tip: apply the stated rule directly to the last given term to verify the answer.

After (10.5), dividing by (2) gives (5.25). Exam tip: identify the next operation in alternating-operation sequences.

The rule is: next term = \(3\times\) previous term \(+1\). Therefore, the term after \(121\) is \(3\times121+1=363+1=364\). Option 363 is only three times 121 and misses the added 1. Exam tip: apply the stated rule directly to the last given term.

In \((7,7,7,\ldots)\), the common difference is \(0\) and the common ratio is \(1\), so it is both an AP and a GP. Option B has a fixed ratio but no fixed difference. Exam tip: check both conditions separately.

The consecutive differences are \(10-7=3\), \(16-10=6\), \(28-16=12\), and \(52-28=24\). Each difference is twice the preceding one, so the next difference is \(48\). Therefore, the next term is \(52+48=100\). Option 96 would require a next difference of 44, which does not follow the doubling rule. Exam tip: For such sequences, first list consecutive differences and check their pattern.

It adds consecutive numbers starting from (1), so (a_n=1+\frac{n(n+1)}{2}). Thus (a_{20}=1+\frac{20\times21}{2}=211).

(6+7(n-1)=83) gives (7(n-1)=77) and (n=12). Exam tip: use the arithmetic sequence formula directly.

Each term is multiplied by (-2), so the eighth term is ((-2)^7=-128). Exam tip: track both sign and power.

The general term is (a_n=5(-2)^{n-1}), so (a_7=5(-2)^6=320). Exam tip: an even power gives a positive sign.

The next terms are (13,21,34,55), so the (10)th term is (55). Exam tip: write positions when extending Fibonacci-type sequences.

In \(7,7,7,7,\ldots\), each consecutive difference is \(7-7=0\), so it is an AP. Each consecutive ratio is \(7/7=1\), so it is also a GP. \(2,5,8,\ldots\) is only an AP. Exam tip: for a constant sequence, check difference 0 and ratio 1.

Given \(a_n=3n^2\), \(a_8=3\times 8^2=192\) and \(a_6=3\times 6^2=108\). Hence, \(a_8-a_6=192-108=84\). Option 78 is incorrect because the values must be found by squaring the term numbers first. Exam tip: substitute each required value of \(n\) into the general term before subtracting.

The differences between consecutive terms are 10−4=6, 18−10=8, 28−18=10, and 40−28=12. These differences increase by 2 each time, so the next difference is 14. Hence, the next term is 40+14=54. Choosing 52 would give a difference of only 12 and would break the pattern of increasing differences. Exam tip: Write the consecutive differences first to identify the pattern in a sequence.

The next difference is (23), so (56+23=79). Exam tip: identifying the difference of differences is the fastest method here.

A quadratic sequence usually has the form \(an^2+bn+c\). Its first differences change, but the second difference is constant and equals \(2a\). An arithmetic sequence has a constant first difference instead. Exam tip: make a difference table to identify it quickly.

The terms are of the form (n^2+1), so the next term is (6^2+1=37). Exam tip: recognize patterns built around squares.