Finding the $n$th term of an AP

Here (a=100), (d=50), so (a_8=100+7\times50=450). The question asks only the (8)th month's deposit, not the total.

Here (a=15), (d=15), so (a_{10}=15+9\times15=150). Treat the equal increase as the common difference.

In this AP, the first term is \(a=21\) and the common difference is \(d=28-21=7\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{14}=21+(14-1)\times7=21+91=112\). Hence, \(112\) is correct. \(119\) is the next, or 15th, term. Exam tip: for the \(n\)th term, add the common difference \(n-1\) times.

Here (a=2.5), (d=2.5), so (a_9=2.5+8\times2.5=22.5). The same formula applies to decimals.

Here (a=\frac{1}{2}), (d=\frac{1}{2}), so (a_{16}=\frac{1}{2}+15\times\frac{1}{2}=8). In fractions, keep denominators clear.

The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{26}=0+(26-1)\times4=25\times4=100\). Hence, the correct answer is \(100\). \(104\) would result from incorrectly adding 26 common differences. Exam tip: always use \((n-1)\) in the formula for the \(n\)th term.

Here (a=-3), (d=-4), so (a_{12}=-3+11(-4)=-47). Check signs twice when both values are negative.

For this AP, the first term is \(a=45\) and the common difference is \(d=39-45=-6\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Thus, \(a_{10}=45+(10-1)(-6)=45-54=-9\). Therefore, \(-9\) is correct. Choosing \(-3\) results from using the number of differences incorrectly. Exam tip: for the \(n\)th term, use \(n-1\) differences, not \(n\).

The nth term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{40}=17+(40-1)\times1=17+39=56\). Hence, 56 is correct. Getting 55 is an off-by-one error, because 39 common differences are added after the first term. Exam tip: always use \(n-1\) when finding the nth term.

Here (a=16), (d=8), so (a_{18}=16+17\times8=152). Do not forget to include the first term.

The general term is \(a_n=4n+6\). Substituting \(n=25\), we get \(a_{25}=4\times25+6=100+6=106\). Hence, 106 is correct. The value 100 is only \(4\times25\); it misses the constant term 6. Exam tip: while finding a term, substitute the given term number carefully for \(n\).

The nth term of an AP is \(a_n=a+(n-1)d\). Thus, \(a_{20}=5+(20-1)(-1)=5-19=-14\). Therefore, \(-14\) is correct. The value \(-13\) would result from incorrectly using 18 in place of \(n-1\). Exam tip: always multiply the common difference by \(n-1\) when finding the nth term.

Here (a=1), (d=2), so (a_{51}=1+50\times2=101). Even for odd numbers, the formula is the safest method.

Here (a=12), (d=4), so (a_9=12+8\times4=44). In time-based questions, the first week is the first term.

For this AP, the first term is \(a=200\) and the common difference is \(d=180-200=-20\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{11}=200+(11-1)(-20)=200-200=0\). The value 20 is the 10th term, not the 11th term. Exam tip: use \(n-1\), not \(n\), in the nth-term formula.

The formula for the nth term of an AP is \(a_n=a+(n-1)d\). Here, \(a=5\), \(d=3\), and \(n=8\), so \(a_8=5+(8-1)\times3=5+21=26\). Therefore, 26 is correct. Choosing 23 results from an incorrect count of differences: there are 7 common differences after the first term to reach the 8th term. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

For this AP, the first term is \(a=2\) and the common difference is \(d=6-2=4\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{12}=2+(12-1)\times4=2+44=46\). Hence, \(46\) is correct. \(48\) would result from incorrectly using \(12d\) instead of \((12-1)d\). Exam tip: always use \((n-1)\) in the formula for the \(n\)th term.

The first term is \(a=7\), and the common difference is \(d=11-7=4\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{10}=7+(10-1)\times4=7+36=43\). Hence, \(43\) is correct. Choosing \(41\) would result from using an incorrect multiplier instead of counting the nine jumps from the first term. Exam tip: always use \((n-1)\) for the \(n\)th term of an AP.

The formula for the nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Thus, \(a_6=12+(6-1)\times5=12+25=37\). Hence, 37 is correct. The answer 42 results from incorrectly using \(n\) instead of \(n-1\). Exam tip: write \(n-1\) separately before substituting the values.

In this AP, the first term is 20 and the common difference is \(d=18-20=-2\). Therefore, \(a_9=a+(9-1)d=20+8(-2)=4\). Hence, 4 is correct. The number 6 is the eighth term, not the ninth term. Exam tip: Use \(n-1\) when finding the \(n\)th term.