Finding the $n$th term of an AP

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.

The direct answer is option B, 45.5 km. The stop distances form an arithmetic progression: the first distance is a=3 km and each next stop is 2.5 km farther, so d=2.5 km. The 18th stop is reached after 17 increases, not 18, because the first stop is already the first term. Hence a_18=3+(18-1)(2.5)=3+17(2.5)=3+42.5=45.5 km. Option A, 43.5 km, is smaller by 2 km and does not result from the formula. Option B, 45.5 km, matches the calculation. Option C, 47.5 km, would come from adding 18 increases instead of 17. Option D, 49.5 km, is still farther and cannot be the 18th value. The word 'first stop' gives the first term, while 'each next stop' gives the common difference. Memory cue: for the kth item, count k-1 moves after the first item.

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.

Apply the nth-term formula a_n = a + (n − 1)d. The first term is a = 55 and the common difference is d = 51 − 55 = −4. Since the nth term is −17, write −17 = 55 + (n − 1)(−4). Subtracting 55 gives −72 = −4(n − 1), and dividing by −4 gives 18 = n − 1. Hence n = 19, so option C is correct. The sign of the common difference must remain negative because the AP decreases by 4. A useful check is that the 19th term is 55 − 18 × 4 = 55 − 72 = −17. Option B would give −13, while option D would give −21, confirming that only 19 works.

The direct answer is D, 112. First find the common difference using the two known terms: \(a_{16}-a_9=77-42=35\), and there are \(16-9=7\) steps, so \(d=35/7=5\). The 23rd term is seven steps after the 16th term. Therefore \(a_{23}=77+7(5)=77+35=112\). Option A, 105, adds only 28 and is one step short. Option B, 108, does not follow the common difference of 5. Option C, 110, is also not obtained by seven increases of 5. Option D, 112, is correct. The important idea is that equally spaced terms in an AP have equal increases: from position 9 to 16 and from 16 to 23, both movements contain seven steps. Count steps between indices rather than counting terms themselves.

The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{17}=9+(17-1)\times\frac{5}{2}=9+16\times\frac{5}{2}=9+40=49\). Hence, the correct answer is 49. Getting 47 would result from using the number of differences incorrectly. Exam tip: for the \(n\)th term, use \(n-1\) common differences, not \(n\).

Given \(a_n=7-2n\), set the nth term equal to \(-35\): \(-35=7-2n\). Thus, \(-42=-2n\), so \(n=21\). Hence, \(-35\) is the 21st term of the AP. The 20th term is \(7-2(20)=-33\), so option C is not correct. Exam tip: To find the term number for a given value, equate that value to \(a_n\) and solve for \(n\).

The governing concept is the constant difference between AP terms. From the 11th to the 19th term there are 19 − 11 = 8 equal steps, while the value increases from 40 to 72, a change of 32. Hence 8d = 32 and d = 4. To move backward from the 11th term to the 4th term requires 11 − 4 = 7 steps, so a₄ = a₁₁ − 7d = 40 − 7 × 4 = 40 − 28 = 12. Therefore option C is correct. The subtraction is necessary because we move backward. Using six or eight steps would produce a wrong distractor, while the correct value must also reproduce the supplied terms: a₁₁ = 12 + 7×4 = 40 and a₁₉ = 12 + 15×4 = 72.

Here, the first term is \(a=17\) and the common difference is \(d=24-17=7\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Thus, \(136=17+(n-1)\times7\), so \(119=7(n-1)\). Hence, \(n-1=17\) and \(n=18\). Option 17 is incorrect because the 17th term is \(129\). Exam tip: always use \((n-1)\), not \(n\), in the AP nth-term formula.

The ((m+5))th term is (5d) ahead of the (m)th term, so (2m+3+10=2m+13). In symbolic terms, look at the position gap.

Given \(a_4=15\) and \(a_{12}=55\). Thus, \(a_{12}-a_4=8d\), so \(55-15=8d\) and \(d=5\). Now \(a_{16}=a_{12}+4d=55+4\times5=75\). Therefore, 75 is correct. The nearby option 80 would result from adding \(5d\) instead of the required \(4d\). Exam tip: use the difference between the term numbers to determine how many common differences are involved.

This is an nth-term application involving a sequence of multiples. The first term is a₁ = 304 because 300 is not greater than 300, and the first multiple of 8 after it is 304. The common difference is d = 312 − 304 = 8. Using aₙ = a₁ + (n − 1)d, the 25th term is a₂₅ = 304 + (25 − 1) × 8 = 304 + 24 × 8 = 304 + 192 = 496. Hence option A is correct. The strict phrase “greater than 300” rules out 300 as a possible starting value, although 300 is itself not a multiple of 8. Taking 25 rather than 24 steps would produce 504, while changing the starting term or making an arithmetic error can lead to the other distractors.

In (11n<500), the greatest (n=45), so the term is (11\times45=495). Take the greatest integer below the limit.

The given formula is \(a_n=7n+12\). For the \(41\)st term, substitute \(n=41\): \(a_{41}=7\times41+12=287+12=299\). Therefore, the correct answer is \(299\). A choice such as \(294\) may result from an arithmetic error. Exam tip: substitute the given value of \(n\) directly into the formula for \(a_n\).

For an arithmetic progression, the difference between every pair of consecutive terms is constant. The condition aₙ₊₁ − aₙ = 9 therefore gives the common difference d = 9. The known term is a₆ = 48, and the required term is a₂₀. From the sixth position to the twentieth position there are 20 − 6 = 14 equal intervals. Therefore, using aₘ = aₖ + (m − k)d, a₂₀ = 48 + (20 − 6) × 9 = 48 + 14 × 9 = 48 + 126 = 174. Hence option C is correct. Adding only 13 differences would miss one transition, while treating 48 as a₁ would disregard the given index and solve a different problem.