Class 10 Maths · Chapter 5
Short answer: Is chapter me 3 exercises hain — Exercise 5.1 (AP identify karna aur first terms/common difference nikalna), Exercise 5.2 (nth term formula ke saare applications — missing terms, kaunsa term hai, do AP ka comparison, salary/word problems), aur Exercise 5.3 (sum of first n terms formula, stadium seating, logs stacking, potato race jaise real-life word problems). Total mila ke ~28 questions cover hote hain, jisme se har type — direct formula, reverse (n ya d nikalna), aur applied word problem — aata hai.
AP me pehchan yahi hai ki consecutive terms ka difference hamesha same rehta hai — usi ko common difference (d) kehte hain. Jaise fare meter, ya ek seedhi seat-row arrangement, ya har hafte fixed amount se badhti saving — jahan bhi "same amount se badhna/ghatna" ka pattern dikhe, wahan AP hai. Ye chapter do formulas ke around ghoomta hai: nth term (an = a + (n−1)d) aur sum of n terms (Sn = n/2[2a + (n−1)d]) — dono ko sahi jagah use karna hi asli skill hai.
Chapter 5 Summary — 5 Minute Revision
1. Arithmetic Progression (AP) kya hai
Ek list of numbers jisme har term (pehle ke alawa) apne pichle term me ek FIXED number jodkar banta hai, use Arithmetic Progression kehte hain. Ye fixed number common difference (d) kehlata hai.
d = a2 − a1 = a3 − a2 = an − an−1
d positive ho sakta hai (increasing AP), negative ho sakta hai (decreasing AP), ya zero bhi ho sakta hai (constant AP, sab terms same).
Check karne ka tarika: kam se kam 3 consecutive terms ka difference nikalo — agar sab EQUAL hain tabhi AP hai. Sirf 2 terms dekh kar decide mat karo.
2. General form aur nth term
AP ka general form: a, a+d, a+2d, a+3d, ...
an = a + (n − 1)d
Yahan a = first term, d = common difference, n = kaunsa term chahiye. Ye formula sबसे zyada use hone wala formula hai — 4 unknowns (a, d, n, an) me se koi bhi 3 diye ho to 4th nikal sakte ho.
3. Sum of first n terms
Do versions hain — jo bhi data diya ho uske hisaab se use karo:
| Kab use karein | Formula |
|---|---|
| Sirf a aur d pata hai | Sn = n/2 [2a + (n−1)d] |
| First aur last term (l) pata hai | Sn = n/2 [a + l] |
↔ Table ko side me swipe karein
Dono formulas ek hi cheez dete hain kyunki l = an = a + (n−1)d hai — bas jo jaldi calculate ho wo use karo.
4. Word problems ka approach
- Sabse pehle sentence se a (starting value) aur d (fixed increase/decrease) identify karo.
- Poocha kya hai — ek specific term (nth term formula) ya total/sum (sum formula) — decide karo.
- Real-life me n hamesha ek positive integer hona chahiye — agar quadratic equation do values de (jaise logs stacking), to jo value context me valid hai wahi lo, doosri reject karo.
- Units mat bhoolo — ₹, seats, logs, metres — final answer me likho.
Poore Class 10 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q24)
Ex 5.1, Q1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) Taxi fare after each km, jab fare ₹15 pehle km ke liye hai aur usske baad har km ke liye ₹8 hai.
(ii) Amount of air present in a cylinder jab vacuum pump har baar cylinder me maujood air ka 1/4 nikal deta hai.
(iii) Cost of digging a well after every metre, jab pehle metre ki cost ₹150 hai aur har agle metre ke liye ₹50 badhti hai.
(iv) Amount of money in an account jab ₹10,000 par 8% p.a. compound interest lagta hai.
(i) Fares: 15, 23, 31, 39, ... — har baar exactly ₹8 badh raha hai (d = 8 constant). Ye AP hai.
d = 23 − 15 = 8, 31 − 23 = 8 — same
(ii) Air remaining: har step me maujood air ka 1/4 nikalta hai, matlab har baar SAME AMOUNT nahi, balki current quantity ka fraction nikal raha hai — difference constant nahi hoga. Ye AP NAHI hai.
(iii) Cost: 150, 200, 250, 300, ... — har metre ₹50 fix badhta hai (d = 50 constant). Ye AP hai.
(iv) Compound interest se amount: 10000, 10800, 11664, ... — difference (800, 864, ...) badhta jaa raha hai, constant nahi. Ye AP NAHI hai.
Ex 5.1, Q2. Diye gaye a (first term) aur d (common difference) se AP ke pehle 4 terms likho.
(i) a = 10, d = 10
(ii) a = −2, d = 0
(iii) a = 4, d = −3
(iv) a = −1, d = 1/2
(v) a = −1.25, d = −0.25
(i) 10, 20, 30, 40
(ii) −2, −2, −2, −2
(iii) 4, 1, −2, −5
(iv) −1, −1/2, 0, 1/2
(v) −1.25, −1.50, −1.75, −2.00
har term = pichla term + d; bas d baar-baar jodte jao
Ex 5.1, Q3. Diye gaye APs ke liye first term aur common difference likho.
(i) 3, 1, −1, −3, ...
(ii) −5, −1, 3, 7, ...
(iii) 1/3, 5/3, 9/3, 13/3, ...
(iv) 0.6, 1.7, 2.8, 3.9, ...
(i) a = 3, d = 1 − 3 = −2
(ii) a = −5, d = −1 − (−5) = 4
(iii) a = 1/3, d = 5/3 − 1/3 = 4/3
(iv) a = 0.6, d = 1.7 − 0.6 = 1.1
Ex 5.1, Q4. Kaunsi list AP hai? Agar AP hai to common difference nikalo aur do aur terms likho.
(i) 2, 4, 8, 16, ...
(ii) 2, 5/2, 3, 7/2, ...
(iii) −1.2, −3.2, −5.2, −7.2, ...
(iv) −10, −6, −2, 2, ...
(v) 3, 3+√2, 3+2√2, 3+3√2, ...
(i) Differences: 4−2=2, 8−4=4 — same nahi. AP nahi hai.
(ii) d = 5/2 − 2 = 1/2 (3 − 5/2 = 1/2 bhi match). AP hai, d = 1/2. Next two terms: 4, 9/2.
a5 = 3 + 4×(1/2) = 4, a6 = 4 + 1/2 = 9/2
(iii) d = −3.2−(−1.2) = −2 (matches −5.2−(−3.2) = −2). AP hai, d = −2. Next: −9.2, −11.2.
(iv) d = −6−(−10) = 4 (matches −2−(−6)=4). AP hai, d = 4. Next: 6, 10.
(v) d = √2 (constant har jagah). AP hai, d = √2. Next: 3+4√2, 3+5√2.
Ex 5.1, Q5. Check karo ki 0.2, 0.22, 0.222, 0.2222, ... AP hai ya nahi.
Differences dekho:
0.22 − 0.2 = 0.02, 0.222 − 0.22 = 0.002, 0.2222 − 0.222 = 0.0002
Differences constant nahi hain (0.02 ≠ 0.002 ≠ 0.0002) — har baar chhota hota jaa raha hai. Ye AP NAHI hai.
Common mistake: pehli nazar me lagta hai numbers 'ek pattern' me badh rahe hain isliye AP hoga — lekin AP ke liye difference EXACT constant hona zaroori hai, sirf 'pattern dikhna' kaafi nahi.
Ex 5.2, Q1. Table complete karo (AP: a, d, n, an):
(a) a = 7, d = 3, n = 8, an = ?
(b) a = −18, n = 10, an = 0, d = ?
(c) a = −3, d = 4, an = 41, n = ?
(a)
a8 = a + 7d = 7 + 7×3 = 7 + 21 = 28
(b)
0 = −18 + 9d ⟹ 9d = 18 ⟹ d = 2
(c)
41 = −3 + (n−1)×4 ⟹ 44 = 4(n−1) ⟹ n−1 = 11 ⟹ n = 12
Ex 5.2, Q2. AP 10, 7, 4, ... ka 30th term nikalo.
a = 10, d = 7 − 10 = −3
a30 = a + 29d = 10 + 29×(−3) = 10 − 87 = −77
Ex 5.2, Q3. AP 2, __, 26 me missing (middle) term nikalo (3 terms ki AP hai).
Yahan a1 = 2, a3 = 26. Common difference:
d = (a3 − a1)/2 = (26 − 2)/2 = 24/2 = 12
missing term = a1 + d = 2 + 12 = 14
(Cross-check: 14 + 12 = 26 ✓)
Ex 5.2, Q4. AP 3, 8, 13, 18, ... ka kaunsa term 78 hai?
a = 3, d = 5. Maan lo nth term 78 hai:
78 = 3 + (n−1)×5
75 = 5(n−1) ⟹ n − 1 = 15 ⟹ n = 16
78, AP ka 16th term hai.
Ex 5.2, Q5. Kitne terms hain in AP: (i) 7, 13, 19, ..., 205 (ii) 18, 15½, 13, ..., −47
(i) a = 7, d = 6, an = 205
205 = 7 + (n−1)×6 ⟹ 198 = 6(n−1) ⟹ n−1 = 33 ⟹ n = 34
(ii) a = 18, d = 15.5 − 18 = −2.5, an = −47
−47 = 18 + (n−1)×(−2.5) ⟹ −65 = −2.5(n−1) ⟹ n−1 = 26 ⟹ n = 27
Ex 5.2, Q6. Kya −150, AP 11, 8, 5, 2, ... ka term hai?
a = 11, d = 8 − 11 = −3. Maan lo −150 = an:
−150 = 11 + (n−1)×(−3)
−161 = −3(n−1) ⟹ n − 1 = 161/3 = 53.67
n integer nahi aa raha (53.67), aur n hamesha whole number hona chahiye. Isliye −150 is AP ka term NAHI hai.
Common mistake: yahan log n non-integer aane par bhi 'closest integer' round off kar dete hain — galat hai, agar n integer na aaye to seedha 'not a term' likho.
Ex 5.2, Q7. Ek AP ka 11th term 38 hai aur 16th term 73 hai. 31st term nikalo.
a16 − a11 = 5d
73 − 38 = 5d ⟹ 35 = 5d ⟹ d = 7
a = a11 − 10d = 38 − 70 = −32
a31 = a + 30d = −32 + 210 = 178
Ex 5.2, Q8. Ek AP me 50 terms hain, jiska 3rd term 12 hai aur last (50th) term 106 hai. 29th term nikalo.
a50 − a3 = 47d
106 − 12 = 47d ⟹ 94 = 47d ⟹ d = 2
a = a3 − 2d = 12 − 4 = 8
a29 = a + 28d = 8 + 56 = 64
Ex 5.2, Q9. Ek AP ka 3rd term 4 hai aur 9th term −8 hai. Kaunsa term shunya (zero) hoga?
a9 − a3 = 6d
−8 − 4 = 6d ⟹ d = −2
a = a3 − 2d = 4 − (−4) = 8
an = 0 ⟹ 8 + (n−1)(−2) = 0 ⟹ n − 1 = 4 ⟹ n = 5
5th term zero hai.
Ex 5.2, Q10 (word problem — salary). Subba Rao ne 1995 me ₹5000 monthly salary par kaam start kiya, aur har saal ₹200 fix increment mila. Kis saal me unki salary ₹7000 ho gayi?
AP: a = 5000, d = 200. Maan lo nth saal me 7000:
7000 = 5000 + (n−1)×200
2000 = 200(n−1) ⟹ n − 1 = 10 ⟹ n = 11
11th term matlab 10 saal baad, i.e. 1995 + 10 = 2005 me unki salary ₹7000 hui.
Dhyan do: n=11 ka matlab 11th saal hai, jo start year (1995) ke 10 saal baad aata hai — off-by-one se bachna important hai.
Ex 5.3, Q1. AP ka sum nikalo: (i) 2, 7, 12, ..., 10 terms tak (ii) −37, −33, −29, ..., 12 terms tak
(i) a = 2, d = 5, n = 10
S10 = 10/2 [2×2 + 9×5] = 5[4 + 45] = 5×49 = 245
(ii) a = −37, d = 4, n = 12
S12 = 12/2 [2×(−37) + 11×4] = 6[−74 + 44] = 6×(−30) = −180
Ex 5.3, Q2. Nikalo: (i) pehle 1000 positive integers ka sum (ii) pehle 15 multiples of 8 ka sum
(i) 1, 2, 3, ..., 1000 — AP hai a=1, d=1, n=1000
S1000 = n(n+1)/2 = 1000×1001/2 = 500500
(ii) 8, 16, 24, ... — a=8, d=8, n=15
S15 = 15/2 [2×8 + 14×8] = 7.5×[16+112] = 7.5×128 = 960
Ex 5.3, Q3. Ek AP me a = 5, d = 3, an = 50 hai. n aur Sn dono nikalo.
50 = 5 + (n−1)×3 ⟹ 45 = 3(n−1) ⟹ n − 1 = 15 ⟹ n = 16
S16 = n/2 [a + an] = 16/2 [5 + 50] = 8×55 = 440
Common mistake: jab last term (an) already pata ho, log fir bhi (n−1)d wala lamba formula use karte hain — Sn = n/2[a+l] use karo, fast hai.
Ex 5.3, Q4. Ek AP me d = 7, 22nd term (a22) = 149 hai. Pehle 22 terms ka sum nikalo.
a = a22 − 21d = 149 − 21×7 = 149 − 147 = 2
S22 = 22/2 [2×2 + 21×7] = 11[4 + 147] = 11×151 = 1661
Ex 5.3, Q5 (word problem — penalty). Ek construction contract me delay penalty ka rule hai: pehle din ₹200, dusre din ₹250, teesre din ₹300 — matlab har agle din penalty ₹50 se badhti jaati hai. Agar contractor 30 din late kare to total kitna penalty dena hoga?
AP: a = 200, d = 50, n = 30
S30 = 30/2 [2×200 + 29×50] = 15[400 + 1450] = 15×1850 = ₹27,750
Ex 5.3, Q6 (word problem — stadium seating). Ek stadium ki 1st row me 20 seats hain, 2nd row me 22, 3rd row me 24, aur is tarah har row me 2 seats zyada hoti jaati hain. Pehli 15 rows me total kitni seats hongi?
AP: a = 20, d = 2, n = 15
S15 = 15/2 [2×20 + 14×2] = 7.5[40 + 28] = 7.5×68 = 510 seats
Ex 5.3, Q7 (word problem — logs stacking). 200 logs ek tarah stack kiye gaye hain — sabse neeche wali row me 20 logs, uske upar wali row me 19, uske upar 18, is tarah har row me 1 log kam hota jaata hai. Kitni rows hongi, aur top row me kitne logs honge?
AP: a = 20, d = −1, Sn = 200
200 = n/2 [2×20 + (n−1)×(−1)] = n/2 [41 − n]
400 = 41n − n² ⟹ n² − 41n + 400 = 0
discriminant = 41² − 4×400 = 1681 − 1600 = 81, √81 = 9
n = (41 ± 9)/2 ⟹ n = 25 ya n = 16
n = 25 lene par a25 = 20 + 24×(−1) = −4 aata hai — logs ki count negative nahi ho sakti, isliye ye value REJECT karo.
n = 16 rows, top row logs = a + (n−1)d = 20 + 15×(−1) = 5 logs
Common mistake: quadratic ke dono roots ko valid maan lena — hamesha context (yahan: log count negative nahi ho sakti) se ek root reject karna padta hai.
Ex 5.3, Q8 (word problem — potato race). 10 potatoes ek seedhi line me rakhe hain, har potato ke beech 3m ka gap hai, pehla potato bucket se 5m door hai. Ek player bucket se start karke ek-ek potato uthata hai, bucket tak wapas laata hai, aur next potato ke liye jaata hai. 10 potatoes uthane me player total kitni doori bhaagega?
Har potato ke liye 'jao aur wapas aao' doori:
1st potato: 2×5 = 10m, 2nd: 2×8 = 16m, 3rd: 2×11 = 22m, ...
Ye AP hai: a = 10, d = 6, n = 10
S10 = 10/2 [2×10 + 9×6] = 5[20 + 54] = 5×74 = 370 metres
Ex 5.3, Q9 (word problem — savings). Ramkali ne saal ke pehle hafte ₹5 bachaye, aur har agle hafte apni bachat ₹1.75 se badhaayi. Agar nth hafte ki bachat ₹20.75 ho gayi ho, to n nikalo aur us tak ki total bachat (Sn) bhi nikalo.
AP: a = 5, d = 1.75, an = 20.75
20.75 = 5 + (n−1)×1.75 ⟹ 15.75 = 1.75(n−1) ⟹ n − 1 = 9 ⟹ n = 10
S10 = 10/2 [2×5 + 9×1.75] = 5[10 + 15.75] = 5×25.75 = ₹128.75
Important Equations — Ek Nazar Me
| Kya nikalna hai | Formula |
|---|---|
| Common difference | d = a2 − a1 = an − an−1 |
| nth term | an = a + (n − 1)d |
| Sum of first n terms (a, d se) | Sn = n/2 [2a + (n − 1)d] |
| Sum of first n terms (a, last term l se) | Sn = n/2 [a + l] |
| nth term aur sum ka relation | an = Sn − Sn−1 |
| Sum of first n natural numbers (special case: a=1,d=1) | Sn = n(n + 1)/2 |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- n aur (n−1) me confuse hona. an = a + (n−1)d hota hai, a + n×d NAHI. Agar (n−1) ki jagah n use kiya to answer ek term aage/peeche shift ho jaata hai — har baar formula likh kar substitute karo, yaad se mat karo.
- Negative d ka sign error. Jab AP decreasing ho (jaise 11, 8, 5, 2, ...), d negative hota hai (d = −3). Calculation me (n−1)×d karte waqt sign galat laga dena bahut common mistake hai — bracket me poora negative number likho: (n−1)×(−3), na ki (n−1)×3 fir end me minus lagana.
- nth term formula aur sum formula ko swap karna. Agar question 'kaunsa term hai' pooche to an formula chahiye; agar 'total kitna' pooche to Sn formula chahiye. In dono ko mix karke ek hi equation me daal dena galat answer deta hai.
- Bina check kiye kisi bhi sequence ko AP maan lena. Sirf 2 terms ka difference dekh kar AP declare mat karo — kam se kam 3 consecutive differences check karo. Jaise 0.2, 0.22, 0.222, ... dikhne me pattern lagta hai lekin AP nahi hai.
- Quadratic se aaye dono roots ko valid maan lena. Jab n ke liye quadratic equation solve karke 2 values aayein (jaise logs stacking problem me n=25 aur n=16), context check kiye bina dono answer likhna galat hai — jo value negative quantity ya invalid scenario deti hai use reject karo.
- Sum formula me a aur l (last term) ko mix karna. Sn = n/2[2a+(n−1)d] aur Sn = n/2[a+l] — dono alag formulas hain. Agar last term (l) already diya hai to seedha dusra formula use karo; pehle formula me last term ko galat jagah daal dena common exam mistake hai.
Board-Style Important Questions
- 1 mark: AP 3, 1, −1, −3, ... ka common difference nikalo.
- 2 marks: AP 5, 9, 13, ... ka nth term nikalo aur isse 15th term calculate karo.
- 2 marks: AP 7, 13, 19, ..., 205 me kitne terms hain?
- 3 marks: AP 10, 15, 20, ... ke pehle 20 terms ka sum nikalo.
- 5 marks: Ek stadium ki rows me seats is tarah arranged hain ki pehli row me 20 seats hain aur har agli row me 2 seats zyada hain. Pehli 15 rows me total seats nikalo. (Ya isi tarah ka koi bhi real-life AP application word problem.)
Quick Quiz — Score Check Karein
Q1. AP me common difference (d) kaise define hota hai?
Q2. Taxi fare: pehle km ke liye ₹15, uske baad har km ke liye ₹8. Ye sequence AP hai kya?
Q3. Cylinder me air ki quantity jab vacuum pump har baar maujood air ka 1/4 nikal deta hai — ye AP kyun NAHI hai?
Q4. AP: a = 4, d = −3 ke pehle 4 terms kya honge?
Q5. AP 10, 7, 4, ... ka 30th term (a30) kya hoga?
Q6. 0.2, 0.22, 0.222, 0.2222, ... AP hai ya nahi?
Q7. Ek AP me 11th term 38 hai aur 16th term 73 hai. 31st term kya hoga?
Q8. Kya −150, AP 11, 8, 5, 2, ... ka koi term hai?
Q9. Sum of first n terms ka formula, jab first term (a) aur last term (l) dono pata ho, kya hai?
Q10. Logs stacking problem: 200 logs stack karne me sabse neeche row me 20 logs, phir har row me 1 kam. Solving se n = 25 aur n = 16 dono aate hain, lekin n = 25 REJECT kyun kiya jaata hai?
Aksar Poochhe Jaane Wale Sawaal
AP aur GP me kya fark hai?
AP me consecutive terms ka DIFFERENCE constant rehta hai (jodna/ghatana), jabki GP me consecutive terms ka RATIO constant rehta hai (multiply/divide karna). 2,4,6,8 AP hai (d=2); 2,4,8,16 GP hai (ratio=2).
Common difference negative ho sakta hai kya?
Haan bilkul. Agar AP decreasing hai (jaise 11, 8, 5, 2, ...) to d negative hoga — is example me d = −3. Formula same rehta hai, bas d ki value negative daalni padti hai.
Kya AP ke terms hamesha whole numbers hone chahiye?
Nahi. AP ke terms fractions, decimals, ya surds (jaise √2 wale) bhi ho sakte hain — jaise 3, 3+√2, 3+2√2, ... bhi ek valid AP hai. Sirf difference constant hona chahiye, terms ka type matter nahi karta.
Agar last term (l) pata na ho to sum kaise nikalein?
Tab Sn = n/2 [2a + (n−1)d] wala formula use karo — isme sirf first term aur common difference chahiye hote hain, last term ki zaroorat nahi padti.
nth term formula ka use kab karte hain, sum formula ka kab?
Jab question 'kaunsa specific term hai' ya 'ek particular position par value kya hai' pooche, tab nth term formula use karo. Jab question 'total kitna' ya 'sabka jod' pooche, tab sum formula use karo.
Real life me AP kahan-kahan use hota hai?
Fixed increment wali salary, fixed rate se badhta rent, stadium/auditorium ki seating rows, logs ya bricks ka stacking pattern, EMI ke through fixed savings, taxi/auto fare structure — jahan bhi 'same fix amount se badhna/ghatna' ho, wahan AP model fit baithta hai.
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