Finding the $n$th term of an AP

The given formula is \(a_n=6n+5\). Putting \(n=32\), \(a_{32}=6\times32+5=192+5=197\). Hence, 197 is correct. The value 191 would result from incorrectly using \(n=31\). Exam tip: Substitute the required term number directly for \(n\) in the formula.

Here (d=8), so (a_{16}=a_4+12d=30+96=126). (a_{n+1}-a_n) is the common difference of the AP.

(a_1+a_2=a+(a+4)=18), so (a=7). Now (a_{10}=7+9\times4=43). First find the first term, then the nth term.

In an AP, the average of equally spaced terms is the middle term, so (a_7=\frac{60}{2}=30). For symmetric terms, the average method is faster.

The terms (64,58,52,\ldots) have (d=-6). The last term greater than (10) is (16) because the next term is (10).

(a=4) and (d=7) so (a_{28}=4+27\times7=193). In exams remember to subtract (1) from the term number.

Here (a=38) and (d=-5) so (a_{17}=38+16(-5)=-42). In a decreasing AP take (d) as negative.

The formula for the nth term of an AP is \(a_n=a+(n-1)d\). Substituting the given values gives \(121=13+(n-1)\times6\). Thus, \(108=6(n-1)\), so \(n-1=18\) and \(n=19\). Option 18 is the value of \(n-1\), not of n. Exam tip: always add 1 after finding \(n-1\).

Comparing with \(a_n=a+(n-1)d\), we get \(a=7\) and \(d=-3\). Option B misses the negative sign. Exam tip: identify the coefficient of \(n-1\) as \(d\).

From (149=21+(n-1)8), (128=8(n-1)) so (n=17). When term number is asked, treat the given term as (a_n).

Here (a=-15) and (d=6) so (a_{24}=-15+23\times6=123). Keep the negative first term in mind.

For an AP, \(a_{14}-a_6=(14-6)d\). Thus, \(79-31=8d\), so \(48=8d\) and \(d=6\). If the common difference were 4, the difference over 8 positions would be only 32, not the given 48. Exam tip: when two non-consecutive terms are given, divide the difference of the terms by the difference of their indices.

(d=\frac{67-27}{13-5}=5) and (a_{21}=67+8\times5=107). Moving from the nearer known term is faster.

For an arithmetic progression, the nth term is a_n=a_1+(n-1)d. In the sequence 96, 87, 78, ..., the common difference is d=87-96=-9. With a_1=96 and n=12, a_12=96+(12-1)(-9)=96-99=-3. Thus option A is correct. The negative difference is essential because the sequence decreases by 9 at every step. A sign error would make the terms increase and could lead to a positive distractor. The calculation also uses 11 differences, not 12, because the first term is already counted as the starting term.

For an AP, the nth-term formula is \(a_n=a+(n-1)d\). Here, \(93=25+(18-1)d\), so \(93=25+17d\). Thus, \(68=17d\) and \(d=4\). If the common difference were 3, the 18th term would be \(25+17\times3=76\), not 93. Exam tip: use \(n-1\), not n, in the nth-term formula.

From a known term of an AP, use \(a_n=a_m+(n-m)d\). Thus, \(a_{18}=a_4+(18-4)d=19+14\times7=19+98=117\). Therefore, 117 is correct. Choosing 119 would result from using an incorrect number of common differences. Exam tip: first find the difference between the term indices, then multiply it by \(d\).

Here (a=1.2) and (d=1.5) so (a_{19}=1.2+18\times1.5=28.2). Multiply decimals carefully.

Given \(a_n=5n+8\). To find the 26th term, substitute \(n=26\): \(a_{26}=5\times26+8=130+8=138\). Therefore, 138 is correct. The value 136 would result from incorrectly adding 6 instead of 8. Exam tip: substitute the term number carefully for \(n\) in the direct formula.

The given direct formula is \(a_n=82-4n\). Substituting \(n=23\), we get \(a_{23}=82-4(23)=82-92=-10\). Therefore, the correct answer is \(-10\). A value such as \(-8\) can result from an incorrect multiplication or subtraction. Exam tip: In a direct formula, substitute the term number for \(n\), multiply first, and then subtract.

Here (a=\frac{3}{4}) and (d=\frac{1}{2}) so (a_{31}=\frac{3}{4}+30\cdot\frac{1}{2}=\frac{63}{4}). Use a common denominator for fractions.