Finding the $n$th term of an AP

The governing formula for the nth term of an arithmetic progression is a_n=a_1+(n−1)d. Here a_10=62, n=10, and d=5. Substitution gives 62=a_1+(10−1)×5=a_1+45. Rearranging, a_1=62−45=17, so option D is correct. The reason for using nine differences is that the first term becomes the tenth term after nine equal steps. A check confirms that starting with 17 and adding 5 nine times gives 17+45=62. If the first term were 12, 15, or 20, the tenth terms would be 57, 60, or 65 respectively, so options A, B, and C fail the given condition.

From (185=14+(n-1)9), (171=9(n-1)) and (n=20). Divide the difference by (d) to get the term number.

(a=-32) and (d=7) so (a_{22}=-32+21\times7=115). Be careful while adding from a negative start.

(d=\frac{104-44}{10}=6) so (a_{28}=104+10\times6=164). Add the same term difference for equal position gaps.

The nth-term formula of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{26}=16+25d\). Given \(a_{26}=116\), we get \(116=16+25d\), so \(25d=100\) and hence \(d=4\). If 3 were used, the 26th term would be \(16+25\times3=91\), not 116. Exam tip: in \(a_n\), the common difference is multiplied by \(n-1\), not by \(n\).

Putting \(n=1\) gives \(a_1=4(1)-7=-3\). The coefficient of \(n\) is 4, so each successive term increases by 4; hence the common difference is 4. Exam tip: always verify the first term using \(n=1\).

Here, the first term is 5 and the common difference is 7. The nth term is \(a_n=5+7(n-1)=7n-2\). From \(7n-2<150\), we get \(n<\frac{152}{7}\), so the greatest integer value of \(n\) is 21. Hence, \(a_{21}=7(21)-2=145\). The numbers 146, 147, and 148 are not terms of this AP because they do not occur when starting from 5 and adding 7 each time. Exam tip: for “less than,” use a strict inequality, not equality.

Putting (n=1), (a_1=1) and the coefficient of (n) is (d=3). The direct formula gives both values quickly.

The 15th row is the 15th term of the AP. \(a_{15}=a+(15-1)d=18+14\times3=60\). Ananya forgot to use \(n-1\). Exam tip: count the increases after the first term, not the row number itself.

Here, the first term is 2 and the common difference is 4. Thus, \(a_n=2+4(n-1)=4n-2\). From \(4n-2<75\), we get \(n<19.25\), so the greatest integer value of n is 19. Therefore, \(a_{19}=74\), which is the last term less than 75. The next term, 76, is not less than 75. Exam tip: For a “last term less than” question, solve the inequality and take the greatest possible integer value of n.

In an AP, each step backwards decreases the term by 6. From the 16th term to the 3rd term, we move back 13 places. Thus, \(a_3=a_{16}-(16-3)d=91-13\times6=91-78=13\). Therefore, the correct answer is 13. A value such as 19 can result from using an incorrect difference between the term numbers. Exam tip: always use the difference of the term indices to find the number of common-difference steps.

(a=3) and (d=2.5) so (a_{18}=3+17\times2.5=45.5) km. In word problems, the first value is (a).

The marks form an arithmetic progression with first term \(a=5\) and common difference \(d=3\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{20}=5+(20-1)\times3=5+57=62\). Hence, the 20th question carries 62 marks. The option 65 results from incorrectly using \(20\) instead of \(20-1\). Exam tip: In the nth-term formula of an AP, always use \((n-1)\).

In an AP, the difference between consecutive terms is constant. For \(a_n=4n-1\), \(a_{n+1}-a_n=4\), so it is an AP. In \(n^2+1\), the differences change. Exam tip: a linear expression in \(n\) represents an AP.

The water amounts form a decreasing AP with first term \(a=180\) and common difference \(d=-12\). The 11th term is \(a_{11}=a+(11-1)d=180+10(-12)=60\). Therefore, the correct answer is \(60\) L. \(48\) L results from subtracting \(12\) eleven times, whereas only ten decreases occur from the first term to the 11th term. Exam tip: always use \(n-1\) in the formula for the \(n\)th term.

This forms an arithmetic progression with first term 350 and common difference 15. The production on the 22nd day is \(a_{22}=a+(22-1)d=350+21\times15=665\). Getting 680 would mean adding 22 differences, but there are only 21 increases from day 1 to day 22. Exam tip: use \(a_n=a+(n-1)d\), paying close attention to \(n-1\).

Since the discount rises by ₹35 each month, the fee decreases by ₹35 each month. Thus, the AP has first term ₹900 and common difference \(d=-35\). Therefore, \(a_{10}=900+(10-1)(-35)=900-315=₹585\). ₹550 would result from subtracting ₹35 ten times, but there are only 9 intervals from the first month to the tenth month. Exam tip: use \(a_n=a+(n-1)d\) for the nth term.

The position gap is (15-9=6) and (d=11), so the difference is (6\times11=66). You do not need to find both terms separately.

For an AP, \(a_n=a+(n-1)d\). Thus, \(a_{20}-a_{12}=8d=94-54=40\), so \(d=5\). Using \(a_{12}=a+11d\), we get \(54=a+11\times5\), hence \(a=-1\). The option 0 would result from subtracting only 10 common differences from \(a_{12}\), but the 12th term is 11 common differences after the first term. Exam tip: always use \((n-1)d\) for the \(n\)th term.

For this AP, the first term is \(a=y\) and the common difference is \(d=6\). The 12th term is \(a_{12}=a+(12-1)d\). Therefore, \(89=y+11\times6=y+66\), so \(y=23\). If \(y=21\), the 12th term would be \(87\), not \(89\). Exam tip: use \(n-1\), not \(n\), in the nth-term formula.