Chapter 3 – Arithmetic Progression (A.P.)
Format: Questions in Red • Answers in Green • Mobile-friendly MathJax.
🚩 20 Most-Important 1-Mark Questions & Solutions
Q1. Define an Arithmetic Progression (A.P.).
Ans. A sequence with constant difference \(d=t_{n+1}-t_n\). Example: \(a,a+d,a+2d,\dots\).
Q2. Write the \(n^{\text{th}}\) term of an A.P. with first term \(a\) and common difference \(d\).
Ans. \(t_n=a+(n-1)d\).
Q3. Write the sum of first \(n\) terms \(S_n\) of an A.P.
Ans. \(S_n=\dfrac{n}{2}\,[2a+(n-1)d]=\dfrac{n}{2}(t_1+t_n)\).
Q4. In \(5,12,19,26,\dots\) find \(d\).
Ans. \(d=7\).
Q5. For \(2,-2,-6,-10,\dots\) find next two terms.
Ans. \(-14,-18\) (\(d=-4\)).
Q6. Is \(1,1,2,2,3,3,\dots\) an A.P.?
Ans. No; differences are not constant.
Q7. If \(a=8,d=-5\), write first four terms.
Ans. \(8,3,-2,-7\).
Q8. In A.P. \(a=3,d=5\). Find \(t_{30}\).
Ans. \(t_{30}=3+29\cdot5=148\).
Q9. Check if \(61\) is a term of \(5,8,11,\dots\).
Ans. Solve \(61=5+(n-1)3\Rightarrow n=\tfrac{59}{3}\notin\mathbb{N}\). No.
Q10. How many two-digit numbers are divisible by \(4\)?
Ans. Terms: \(12,16,\dots,96\Rightarrow n=\dfrac{96-12}{4}+1=22\).
Q11. If \(t_{10}=25,\ t_{18}=41\) in an A.P., find \(d\).
Ans. \(t_{18}-t_{10}=8d=16\Rightarrow d=2\).
Q12. For above, find \(a\).
Ans. \(a=t_{10}-9d=25-18=7\).
Q13. If \(t_n=99\) for the A.P. with \(a=7,d=2\), find \(n\).
Ans. \(7+(n-1)2=99\Rightarrow n=47\).
Q14. Sum of first \(n\) natural numbers.
Ans. \(S_n=\dfrac{n(n+1)}{2}\).
Q15. Sum of first \(n\) even natural numbers.
Ans. \(2+4+\dots+2n=n(n+1)\).
Q16. Sum of first \(n\) odd natural numbers.
Ans. \(1+3+\dots+(2n-1)=n^2\).
Q17. For \(3/2,1/2,-1/2,\dots\), find \(d\).
Ans. \(d=-1\).
Q18. Decide A.P.: \(0.3,0.33,0.0333,\dots\)
Ans. Not A.P. (differences not constant).
Q19. First term and \(d\) of \(5,1,-3,-7,\dots\).
Ans. \(a=5,\ d=-4\).
Q20. If \(a=1,d=0,n=101\), find \(t_n\).
Ans. Constant sequence ⇒ \(t_{101}=1\).
📝 20 Most-Important 2-Mark Questions & Solutions
Q1. Which of these are A.P.? (i) \(2,4,6,8,\dots\) (ii) \(2,\tfrac52,3,\tfrac73,\dots\) (iii) \(-10,-6,-2,2,\dots\).
Ans. (i) A.P., \(d=2\). (ii) Not A.P. (differences change). (iii) A.P., \(d=4\).
Q2. Write the A.P. for \(a=10,d=5\) (first 4 terms).
Ans. \(10,15,20,25,\dots\).
Q3. For \(127,135,143,151,\dots\) find \(a,d\) and \(t_{10}\).
Ans. \(a=127,d=8,\ t_{10}=127+9\cdot8=199\).
Q4. Find the \(24^{\text{th}}\) term of \(12,16,20,24,\dots\).
Ans. \(a=12,d=4\Rightarrow t_{24}=12+23\cdot4=104\).
Q5. Find the \(19^{\text{th}}\) term of \(7,13,19,25,\dots\).
Ans. \(a=7,d=6\Rightarrow t_{19}=7+18\cdot6=115\).
Q6. Find the \(27^{\text{th}}\) term of \(9,4,-1,-6,-11,\dots\).
Ans. \(a=9,d=-5\Rightarrow t_{27}=9+26(-5)=-121\).
Q7. How many 3-digit natural numbers are divisible by \(5\)?
Ans. \(100,105,\dots,995\Rightarrow n=\dfrac{995-100}{5}+1=180\).
Q8. In an A.P. \(t_{11}=16,t_{21}=29\). Find \(t_{41}\).
Ans. \(t_{21}-t_{11}=10d=13\Rightarrow d=\tfrac{13}{10}\). \(a=t_{11}-10d=16-13=3\). \(t_{41}=a+40d=3+52=55\).
Q9. In \(11,8,5,2,\dots\) which term is \(-151\)?
Ans. \(a=11,d=-3\). Solve \(11+(n-1)(-3)=-151\Rightarrow -3n=-159\Rightarrow n=53\).
Q10. How many integers from \(10\) to \(250\) are divisible by \(4\)?
Ans. Sequence \(12,16,\dots,248\Rightarrow n=\dfrac{248-12}{4}+1=60\).
Q11. If in an A.P. \(t_{17}-t_{10}=7\), find \(d\).
Ans. \(7d=7\Rightarrow d=1\).
Q12. Find \(t_{63}\) of \(2,11,20,29,\dots\).
Ans. \(a=2,d=9\Rightarrow t_{63}=2+62\cdot9=560\).
Q13. If \(a=200,d=50\), in which month will saving reach \(1000\)?
Ans. \(200+(n-1)50=1000\Rightarrow n=17\).
Q14. Find number of two-digit multiples of \(3\).
Ans. \(12,15,\dots,99\Rightarrow n=\dfrac{99-12}{3}+1=30\).
Q15. If \(t_{10}=a+9d\) and \(t_{18}=a+17d\) are \(25\) and \(41\), find \(t_{38}\).
Ans. From earlier: \(a=7,d=2\Rightarrow t_{38}=7+37\cdot2=81\).
Q16. Sum of first \(100\) terms of \(14,16,18,\dots\).
Ans. \(a=14,d=2,n=100\Rightarrow S_{100}=50[28+198]=11300\).
Q17. Sum of first \(n\) odd numbers equals?
Ans. \(n^2\).
Q18. Number of terms in \(4,8,\dots,136\) and their sum.
Ans. \(n=\dfrac{136-4}{4}+1=34\). \(S_n=\dfrac{34}{2}(4+136)=17\cdot140=2380\).
Q19. If \(S_{55}=3300\) for an A.P., find \(t_{28}\).
Ans. \(t_{28}=\dfrac{S_{28}-S_{27}}{}\). But also \(t_{28}=\dfrac{S_{55}}{55}\cdot2 - a - 27d\) is indirect. Use property \(t_{(n+1)/2}=\dfrac{2S_n}{n}\) for odd \(n\). Here \(n=55\Rightarrow t_{28}=\dfrac{2S_{55}}{55}=\dfrac{6600}{55}=120\).
Q20. If \(t_1=1,\ t_n=20,\ S_n=399\), find \(n\).
Ans. \(S_n=\dfrac{n}{2}(1+20)=\dfrac{21n}{2}=399\Rightarrow n=38\).
🎯 20 Most-Important 3-Mark Questions & Solutions
Q1. For \(a=6,d=3\), find \(S_{27}\).
Ans. \(S_{27}=\dfrac{27}{2}[2\cdot6+(27-1)\cdot3]=\dfrac{27}{2}(12+78)=\dfrac{27}{2}\cdot90=1215\).
Q2. Sum of first \(123\) even natural numbers.
Ans. \(n(n+1)=123\cdot124=15252\).
Q3. Sum of all even numbers between \(1\) and \(350\).
Ans. \(2,4,\dots,350\Rightarrow n=\dfrac{350}{2}=175,\ S=\dfrac{175}{2}(2+350)=175\cdot176=30800\).
Q4. In an A.P. \(t_{19}=52,\ t_{38}=128\). Find \(S_{56}\).
Ans. \(t_{38}-t_{19}=19d=76\Rightarrow d=4\). \(a=t_{19}-18d=52-72=-20\). Then \(S_{56}=\dfrac{56}{2}[2(-20)+(56-1)\cdot4]=28(-40+220)=28\cdot180=5040\).
Q5. Sum of numbers from \(1\) to \(140\) divisible by \(4\).
Ans. A.P. \(4,8,\dots,136\). \(n=34\). \(S=\dfrac{34}{2}(4+136)=17\cdot140=2380\).
Q6. In an A.P., sum of three consecutive terms is \(27\) and their product is \(504\). Find the terms.
Ans. Let \(a-d,a,a+d\). Sum \(=3a=27\Rightarrow a=9\). Product \(=(9-d)\cdot9\cdot(9+d)=9(81-d^2)=504\Rightarrow 81-d^2=56\Rightarrow d^2=25\Rightarrow d=5\). Terms: \(4,9,14\).
Q7. Four consecutive A.P. terms sum to \(12\) and sum of \(3^{\text{rd}}+4^{\text{th}}=14\). Find the four terms.
Ans. Take \(a-d,a,a+d,a+2d\). Total \(=4a+2d=12\) …(i); last two \(=2a+3d=14\) …(ii). Solve: from (i) \(2a+d=6\). From (ii) \(2a+3d=14\Rightarrow 2d=8\Rightarrow d=4\). Then \(2a+d=6\Rightarrow 2a=2\Rightarrow a=1\). Terms: \(-3,1,5,9\).
Q8. If \(t_9=0\) in an A.P., show \(t_{29}=2t_{19}\).
Ans. \(t_n=a+(n-1)d\). \(t_{29}-2t_{19}=[a+28d]-2[a+18d]=a+28d-2a-36d=-(a+8d)=-t_9=0\Rightarrow t_{29}=2t_{19}\).
Q9. Mixer factory: \(t_3=600,\ t_7=700\). Find (i) first year (ii) \(10^{\text{th}}\) year (iii) total of first 7 years.
Ans. \(a+2d=600,\ a+6d=700\Rightarrow d=25,\ a=550\). (i) \(550\). (ii) \(t_{10}=550+9\cdot25=775\). (iii) \(S_7=\dfrac{7}{2}[2\cdot550+6\cdot25]=\dfrac{7}{2}(1100+150)=4375\).
Q10. Loan: total \(3,25,000\), first month \(30500\), each next is \(1500\) less. Find months.
Ans. \(a=30500,d=-1500\). Solve \(S_n=\dfrac{n}{2}[2\cdot30500+(n-1)(-1500)]=325000\Rightarrow (n-20)(3n-65)=0\Rightarrow n=20\) (natural). So \(20\) months.
Q11. Spiral of 13 semicircles with radii \(0.5,1,1.5,\dots\) cm. Find total length (use \(\pi=\tfrac{22}{7}\)).
Ans. Semicircumferences: \(\tfrac{\pi}{2},\pi,\tfrac{3\pi}{2},\dots\) is A.P. in multiples of \(\tfrac{\pi}{2}\) with \(a=\tfrac{\pi}{2},d=\tfrac{\pi}{2}\). \(S_{13}=\dfrac{13}{2}[2a+12d]=\dfrac{13}{2}[\pi+6\pi]=\dfrac{13}{2}\cdot7\pi= \dfrac{91\pi}{2}=143\) cm.
Q12. Literacy: \(2010\) had \(4000\), increase \(400\) yearly. Find 2020.
Ans. \(n=11,\ t_{11}=4000+10\cdot400=8000\).
Q13. Salary: starts \(1{,}80{,}000\), yearly increment \(10{,}000\). In which year is it \(2{,}50{,}000\)?
Ans. \(1.8+0.01(n-1)=2.5\) (in lakhs) ⇒ \(n=8\). So \(8^{\text{th}}\) year.
Q14. Find the \(4^{\text{th}}\) term from the end in \(-11,-8,-5,\dots,49\).
Ans. From end, \(t'_{4}=t_{n-4+1}\) where \(n=\dfrac{49-(-11)}{3}+1=21\). So \(t_{18}= -11+17\cdot3=40\).
Q15. In an A.P. \(t_{10}=46,\ t_5+t_7=52\). Find the A.P.
Ans. \(t_{10}=a+9d=46\). Also \(t_5+t_7=(a+4d)+(a+6d)=2a+10d=52\Rightarrow a+5d=26\). Subtract: \((a+9d)-(a+5d)=20\Rightarrow 4d=20\Rightarrow d=5\). Then \(a=26-25=1\). A.P.: \(1,6,11,\dots\).
Q16. A.P. with \(t_4=-15,\ t_9=-30\). Find \(S_{10}\).
Ans. \(a+3d=-15,\ a+8d=-30\Rightarrow 5d=-15\Rightarrow d=-3,\ a=-6\). \(S_{10}=\dfrac{10}{2}[2(-6)+9(-3)]=5(-12-27)=-195\).
Q17. Two A.P.s: \(9,7,5,\dots\) and \(24,21,18,\dots\). If their \(n^{\text{th}}\) terms are equal, find \(n\) and the term.
Ans. \(9+(n-1)(-2)=24+(n-1)(-3)\Rightarrow 11-2n=27-3n\Rightarrow n=16\). Term \(=9-30=-21\).
Q18. If \(t_3+t_8=7\) and \(t_7+t_{14}=-3\), find \(t_{10}\).
Ans. \( (a+2d)+(a+7d)=2a+9d=7\) …(i). \((a+6d)+(a+13d)=2a+19d=-3\) …(ii). Subtract: \(10d=-10\Rightarrow d=-1\). From (i) \(2a-9=7\Rightarrow a=8\). \(t_{10}=a+9d=8-9=-1\).
Q19. In an A.P. first term \(-5\), last term \(45\), sum \(120\). Find number of terms and \(d\).
Ans. \(\dfrac{n}{2}(a+l)=120\Rightarrow \dfrac{n}{2}(40)=120\Rightarrow n=6\). Then \(d=\dfrac{l-a}{n-1}=\dfrac{45-(-5)}{5}=10\).
Q20. If sum \(1+2+\dots+n=36\), find \(n\).
Ans. \(\dfrac{n(n+1)}{2}=36\Rightarrow n^2+n-72=0\Rightarrow n=8\) (positive root).
📘 Textbook Exercise Questions & Perfect Solutions
Practice Set 3.1
Q1. Which of the following are A.P.? Also find \(d\) if A.P.: (1) \(2,4,6,8,\dots\) (2) \(2,\tfrac52,3,\tfrac73,\dots\) (3) \(-10,-6,-2,2,\dots\) (4) \(0.3,0.33,0.0333,\dots\) (5) \(0,-4,-8,-12,\dots\) (6) \(-\tfrac{1}{5},-\tfrac{1}{5},-\tfrac{1}{5},\dots\) (7) \(3,3+\sqrt2,3+2\sqrt2,3+3\sqrt2,\dots\) (8) \(127,132,137,\dots\).
- (1) A.P., \(d=2\)
- (2) Not A.P.
- (3) A.P., \(d=4\)
- (4) Not A.P.
- (5) A.P., \(d=-4\)
- (6) A.P., \(d=0\)
- (7) A.P., \(d=\sqrt2\)
- (8) A.P., \(d=5\)
Q2. Write A.P. for given \(a,d\): (1) \(10,5\) (2) \(-3,0\) (3) \(-7,\tfrac12\) (4) \(-1.25,3\) (5) \(6,-3\) (6) \(-19,-4\) (first 4 terms).
- (1) \(10,15,20,25\)
- (2) \(-3,-3,-3,-3\)
- (3) \(-7,-6.5,-6,-5.5\)
- (4) \(-1.25,1.75,4.75,7.75\)
- (5) \(6,3,0,-3\)
- (6) \(-19,-23,-27,-31\)
Q3. Find \(a\) and \(d\): (1) \(5,1,-3,-7,\dots\) (2) \(0.6,0.9,1.2,1.5,\dots\) (3) \(127,135,143,151,\dots\) (4) \(\tfrac14,\tfrac34,\tfrac54,\tfrac74,\dots\).
- (1) \(a=5,d=-4\)
- (2) \(a=0.6,d=0.3\)
- (3) \(a=127,d=8\)
- (4) \(a=\tfrac14,d=\tfrac12\)
Practice Set 3.2
Q1. Fill the blanks for each A.P. (i) \(1,8,15,22,\dots\) (ii) \(3,6,9,12,\dots\) (iii) \(-3,-8,-13,-18,\dots\) (iv) \(70,60,50,40,\dots\).
- (i) \(a=t_1=1,\ t_2=8,\ t_3=15,\ d=t_2-t_1=7=t_3-t_2\).
- (ii) \(t_1=3,t_2=6,t_3=9,t_4=12,\ d=3\).
- (iii) \(t_1=-3,t_2=-8,t_3=-13,t_4=-18;\ a=-3,\ d=-5\).
- (iv) \(t_1=70,t_2=60,t_3=50;\ a=70,\ d=-10\).
Q2. Decide if \(-12,-5,2,9,16,23,30,\dots\) is an A.P. If yes, find \(t_{20}\).
Ans. Differences \(=7\) constant ⇒ A.P. \(a=-12,d=7\Rightarrow t_{20}=-12+19\cdot7=121\).
Q3. In \(12,16,20,24,\dots\), find \(t_{24}\).
Ans. \(104\) (as above).
Q4. Find \(t_{19}\) of \(7,13,19,25,\dots\).
Ans. \(115\) (as above).
Q5. Find \(t_{27}\) of \(9,4,-1,-6,-11,\dots\).
Ans. \(-121\).
Q6. How many three-digit natural numbers are divisible by \(5\)?
Ans. \(180\) (as above).
Q7. If \(t_{11}=16,t_{21}=29\), find \(t_{41}\).
Ans. \(55\) (as above).
Q8. \(11,8,5,2,\dots\): which term is \(-151\)?
Ans. \(53^{\text{rd}}\) term.
Q9. Count numbers divisible by \(4\) in \(10\) to \(250\).
Ans. \(60\).
Q10. In an A.P., \(t_{17}\) is \(7\) more than \(t_{10}\). Find \(d\).
Ans. \(d=1\) (since \(t_{17}-t_{10}=7d=7\)).
Practice Set 3.3
Q1. \(a=6,d=3\). Find \(S_{27}\).
Ans. \(1215\) (as above).
Q2. Sum of first \(123\) even natural numbers.
Ans. \(15252\).
Q3. Sum of all even numbers between \(1\) and \(350\).
Ans. \(30800\).
Q4. In an A.P. \(t_{19}=52,\ t_{38}=128\). Find \(S_{56}\).
Ans. \(5040\).
Q5. Sum of numbers \(1\) to \(140\) divisible by \(4\).
Ans. \(2380\).
Q6. If \(S_{55}=3300\), find \(t_{28}\).
Ans. \(t_{28}=\dfrac{2S_{55}}{55}=120\).
Q7. In an A.P., sum of three consecutive terms \(=27\) and product \(=504\). Find terms.
Ans. \(4,9,14\).
Q8. Four consecutive terms have sum \(12\) and \(t_3+t_4=14\). Find them.
Ans. \(-3,1,5,9\).
Q9. If \(t_9=0\), prove \(t_{29}=2t_{19}\).
Ans. Shown above.
Practice Set 3.4
Q1. Sanika saves ₹\(10,11,12,\dots\) daily through 2016. Find total by 31-Dec-2016.
Ans. 2016 is leap year ⇒ 366 days. \(a=10,d=1,n=366\). \(S=\dfrac{366}{2}[2\cdot10+365\cdot1]=183(20+365)=183\cdot385=70455\) rupees.
Q2. Borrowed ₹\(8000\); total interest ₹\(1360\) over 12 monthly installments forming A.P. decreasing by ₹40. Find first and last installment.
Ans. Total repay \(=8000+1360=9360=S_{12}=\dfrac{12}{2}(2a+11d)=6(2a-440)\) since \(d=-40\). So \(6(2a-440)=9360\Rightarrow 2a-440=1560\Rightarrow a=1000\). Last \(=a+11d=1000-440=560\) (rupees).
Q3. Sachin invests ₹\(5000,7000,9000,\dots\) for 12 years. Find total.
Ans. \(a=5000,d=2000,n=12\Rightarrow S=\dfrac{12}{2}[2\cdot5000+11\cdot2000]=6(10000+22000)=6\cdot32000=192000\) rupees.
Q4. Auditorium rows: first row 20 seats, then \(+2\) each row; \(27\) rows. Find seats in \(15^{\text{th}}\) row and total seats.
Ans. \(a=20,d=2\). \(t_{15}=20+14\cdot2=48\). Total \(S_{27}=\dfrac{27}{2}[2\cdot20+26\cdot2]=\dfrac{27}{2}(40+52)=\dfrac{27}{2}\cdot92=27\cdot46=1242\) seats.
Q5. Kargil week temperatures in A.P.; \(M+Sat\) is \(5^\circ\) more than \(Tue+Sat\). If Wed \(=-30^\circ\)C, find others.
Ans. Let Mon \(=a-2d\), Tue \(=a-d\), Wed \(=a\), Thu \(=a+d\), Fri \(=a+2d\), Sat \(=a+3d\) (6 days). Given Wed \(=-30\Rightarrow a=-30\). Condition: \((a-2d)+(a+3d)=(a-d)+(a+3d)+5\Rightarrow a+d=a+2d+5\Rightarrow d=-5\). Thus: Mon \(-20\), Tue \(-25\), Wed \(-30\), Thu \(-35\), Fri \(-40\), Sat \(-45\) (°C).
Q6. Triangular tree rows: \(1,2,3,\dots\) trees per row up to 25 rows. Find total trees.
Ans. \(S_{25}=\dfrac{25\cdot26}{2}=325\) trees.
Applications (Solved Examples Recap)
Q1. Mixer production: \(t_3=600,\ t_7=700\). Find (i) first year (ii) \(10^{\text{th}}\) year (iii) total first 7 years.
Ans. (i) \(550\) (ii) \(775\) (iii) \(4375\) (mixers).
Q2. Decreasing monthly loan payment to clear ₹\(3{,}25{,}000\): first ₹\(30500\), step \(-1500\).
Ans. \(20\) months.
Q3. Saving ₹\(200,250,300,\dots\): in which month ₹\(1000\)? Also total in 17 months.
Ans. \(17^{\text{th}}\) month; total \(S_{17}=10200\) rupees.
Q4. Spiral of 13 semicircles (radii \(0.5,1,1.5,\dots\) cm). Find total length with \(\pi=\frac{22}{7}\).
Ans. \(143\) cm.
Q5. Literate population growth \(+400\) per year: 2010 → 4000. Find 2020.
Ans. \(8000\).
Q6. Salary from \(2015\): ₹\(1{,}80{,}000\) plus ₹\(10{,}000\)/year. When ₹\(2{,}50{,}000\)?
Ans. \(8^{\text{th}}\) year.
Problem Set – 3 (Fully Solved)
Q1. MCQ Keys: (1)–(10).
- (1) (B) \(d=4\)
- (2) (C) \(-2,-4,-6,-8\)
- (3) (B) \(465\) (since \( \frac{30\cdot31}{2}\))
- (4) \(t_7=a+6d=4,\ d=-4\Rightarrow a=28\) ⇒ (D) \(28\)
- (5) (B) \(3.5\) (constant \(d=0\))
- (6) \(a=-3,d=7\Rightarrow t_{21}=-3+20\cdot7=137\) ⇒ (C)
- (7) \(t_{18}-t_{13}=5d\Rightarrow 25\) ⇒ (C)
- (8) First five multiples of 3: \(3,6,9,12,15\Rightarrow 45\) ⇒ (A)
- (9) \(a=15,d=-5,n=10\Rightarrow S_{10}=\frac{10}{2}[30+9(-5)]=5(30-45)=-75\) ⇒ (A)
- (10) \( \frac{n}{2}(1+20)=399\Rightarrow n=38\) ⇒ (B)
Q2. Find the \(4^{\text{th}}\) term from the end in \(-11,-8,-5,\dots,49\).
Ans. Total terms \(n=\dfrac{49-(-11)}{3}+1=21\). \(4^{\text{th}}\) from end \(=t_{21-4+1}=t_{18}=-11+17\cdot3=40\).
Q3. In an A.P., \(t_{10}=46\) and \(t_5+t_7=52\). Find the A.P.
Ans. A.P. \(1,6,11,\dots\) (from 3-Mark Q15).
Q4. For A.P. with \(t_4=-15,\ t_9=-30\), find \(S_{10}\).
Ans. \(-195\) (as above).
Q5. Two A.P.s \(9,7,5,\dots\) and \(24,21,18,\dots\). If \(t_n\) equal, find \(n\) and \(t_n\).
Ans. \(n=16,\ t_n=-21\).
Q6. If \(t_3+t_8=7\) and \(t_7+t_{14}=-3\), find \(t_{10}\).
Ans. \(-1\) (as above).
Q7. First term \(-5\), last \(45\), sum \(120\). Find \(n\) and \(d\).
Ans. \(n=6,\ d=10\) (as above).
Q8. If \(1+2+\dots+n=36\), find \(n\).
Ans. \(n=8\).
🔑 Quick Summary
- nth term: \(t_n=a+(n-1)d\)
- Sum: \(S_n=\dfrac{n}{2}[2a+(n-1)d]=\dfrac{n}{2}(t_1+t_n)\)
- Identify A.P.: check constant difference \(d\)
- Counting terms: \(n=\dfrac{l-a}{d}+1\) (last \(l\))
- Odd/Even sums: \(1+3+\cdots+(2n-1)=n^2\), \(2+4+\cdots+2n=n(n+1)\)