3.3.1 · Maths › Sequences & Series
AP ek aisi sequence hai jahan aap har step pe same amount add karte ho . Woh fixed jump hi ==common difference d == hai. Kyunki har step identical hai, AP ke baare mein sab kuch (koi bhi term, koi bhi sum) sirf do numbers se rebuild ho sakta hai: start a aur step d .
Definition Arithmetic Progression
Ek sequence a 1 , a 2 , a 3 , … ek AP hai agar consecutive terms ka difference constant ho:
a n + 1 − a n = d (same for all n )
First term = a (ise a 1 bhi likhte hain)
Common difference = d = a 2 − a 1 = a 3 − a 2 = …
Example: 2 , 5 , 8 , 11 , … mein a = 2 , d = 3 hai.
WHY ise difference se define karte hain? Kyunki "constant difference" hi woh ek aur ekmaatra rule hai jo AP follow karta hai. Ek baar yeh pata chal gaya, toh har term automatically determined ho jaata hai.
Intuition WHY ek formula hai
Term n tak pahunchne ke liye, aap a se start karte ho aur d size ke steps lete ho. Kitne steps? Term 1 se term n tak jaane ke liye aap ( n − 1 ) baar jump karte ho — n baar nahi , kyunki pehli term ke liye zero jumps lagte hain.
HOW — ise build up karo:
a 1 a 2 a 3 a 4 a n = a = a + d = a + 2 d = a + 3 d ⋮ = a + ( n − 1 ) d ( 0 steps ) ( 1 step ) ( 2 steps ) ( 3 steps ) ( n − 1 steps )
Intuition WHY pairing kaam karta hai
Sum ko aage likhो, phir neeche ulta likhो. Har vertical pair same value mein add hoti hai (first + last), kyunki jaise ek number d se badhta hai, neeche wala d se ghatta hai — woh cancel ho jaate hain. Toh saare n pairs equal hain, aur sum karna sirf multiplication ban jaata hai.
HOW — derivation:
Sum likhо, phir use reverse karo:
S n S n = a + ( a + d ) + ( a + 2 d ) + ⋯ + ( a + ( n − 1 ) d ) = ( a + ( n − 1 ) d ) + ⋯ + ( a + 2 d ) + ( a + d ) + a
Dono lines ko column by column add karo. Har column 2 a + ( n − 1 ) d = a 1 + a n deta hai:
2 S n = n equal terms ( 2 a + ( n − 1 ) d ) + ⋯ + ( 2 a + ( n − 1 ) d ) = n ( 2 a + ( n − 1 ) d )
2 se divide karo:
WHY doosri form beautiful hai: sum = (number of terms) × (average term). AP mein average sirf 2 first + last hota hai kyunki terms evenly spaced hoti hain.
3 , 7 , 11 , … ki 20th term nikalo
a = 3 , d = 7 − 3 = 4 . Yeh step kyun? d koi bhi consecutive difference hota hai.
a 20 = a + ( 20 − 1 ) d = 3 + 19 ⋅ 4 . 19 kyun? Hum n − 1 = 19 baar jump karte hain.
= 3 + 76 = 79 .
Worked example 2 — Pehle 100 natural numbers ka sum
Yahan a = 1 , d = 1 , n = 100 . AP kyun? Difference constant = 1 hai.
Last term a 100 = 1 + ( 99 ) ( 1 ) = 100 .
S 100 = 2 100 ( a + a 100 ) = 50 × 101 = 5050 . Yeh form kyun? Hume first & last pata hai, toh average form sabse fast hai.
5 , 8 , 11 , … ke kitne terms 84 sum karte hain?
a = 5 , d = 3 , S n = 84 .
84 = 2 n [ 2 ( 5 ) + ( n − 1 ) 3 ] = 2 n ( 3 n + 7 ) .
168 = 3 n 2 + 7 n ⇒ 3 n 2 + 7 n − 168 = 0 .
n = 6 − 7 ± 49 + 2016 = 6 − 7 ± 45 . Negative root reject kyun karte hain? n ek positive whole number hona chahiye.
n = 6 38 (reject) ya n = 6 − 52 (reject)… recompute: 2065 ≈ 45.4 , jisse n ≈ 6.4 milta hai — koi integer solution nahi, toh koi exact n in values ke saath 84 nahi deta. (Lesson: hamesha check karo ki discriminant ek whole number deta hai.)
( n − 1 ) steps ki jagah n use karna
Kyun sahi lagta hai: "20th term, 20 of something" — aise lagta hai 20 se multiply karo.
Fix: Pehli term free hai (0 jumps). Sirf n − 1 jumps of d hote hain. Term 1 pe test karo: a + ( 1 − 1 ) d = a ✓.
d = a 2 / a 1 (subtract ki jagah divide karna)
Kyun sahi lagta hai: AP ko GP se confuse karna, jahan ratio ke liye divide karte hain.
Fix: AP = constant add karna ⇒ d = a 2 − a 1 (subtract). GP = multiply karna ⇒ ratio.
n ko positive integer check karna bhool jaana
Kyun sahi lagta hai: Tumne quadratic solve ki, ek number mila — ho gaya!
Fix: n terms count karta hai; negatives, fractions reject karo, aur dono roots check karo.
Recall Ye teen master karo
a n = a + ( n − 1 ) d
S n = 2 n [ 2 a + ( n − 1 ) d ] = 2 n ( a + a n )
a n = S n − S n − 1 ; aur n mein a n linear ⇔ AP .
Recall Feynman: ek 12-year-old ko explain karo
Socho tum ek aisi stairs chadh rahe ho jahan har step ki height same hai . Tumhari starting jagah a hai; har step tumhe d se upar uthati hai. Step number n pe tumhari height jaanne ke liye, tumne sirf ( n − 1 ) steps chadhe hain (step 1 tak pahunchne ke liye tum chadhe nahi — tum already wahan the!), toh height = a + ( n − 1 ) d .
Ab saari stairs ki heights add karne ke liye: unhe shortest se tallest tak line up karo, phir same stairs tallest se shortest tak ek side mein rakh do. Ab har pair ki total height same hai. n pairs hain, har ek = ( first + last ) , toh double answer = n ( first + last ) . Ise halve karo aur ho gaya!
"nth term: step lene se pehle ONE subtract karo" → ( n − 1 ) d .
Sum = "Kitne × Average" → S n = n × 2 first + last .
Geometric Progression (GP) — multiply-by-ratio cousin of the add-by-d AP.
Arithmetic Mean — 3-term AP ka middle term = 2 a + c .
Sigma Notation — S n = ∑ k = 1 n a k .
Quadratic Equations — "kitne terms" wale problems solve karne ke liye use hota hai.
Gauss Summation Trick — S n ke peeche pairing idea.
Linear Functions — a n , n mein ek straight line hai.
AP ko kya define karta hai? Consecutive terms ka difference constant hota hai, a n + 1 − a n = d .
AP ke nth term ka formula? a n = a + ( n − 1 ) d .
( n − 1 ) d kyun, n d kyun nahi?Pehli term ke liye 0 jumps chahiye; term n tak pahunchne mein sirf n − 1 jumps of size d lagte hain.
n terms ka sum (standard form)? S n = 2 n [ 2 a + ( n − 1 ) d ] .
n terms ka sum (average/first-last form)? S n = 2 n ( a + a n ) , yani terms × first aur last ka average.
Partial sums se a n kaise nikalte hain? a n = S n − S n − 1 .
Agar a n , n mein linear hai, toh yeh kya batata hai? Sequence ek AP hai (uska common difference slope hai).
Agar S n , n mein quadratic hai (bina constant term ke), toh sequence kya hai? Ek AP.
2 , 5 , 8 , 11 ka common difference?d = 3 (consecutive terms subtract karo).
S n derive karne ka trick?Sum ko aage + ulta likhо aur add karo; har column a + a n equal hota hai.
S_n equals n over 2 times 2a plus n-1 d
S_n equals n over 2 times a plus a_n
a_n equals S_n minus S_n-1