An AP is a sequence where you add the same amount every step . That fixed jump is the ==common difference d d d start a a a step d d d
Definition Arithmetic ProgressionA sequence a 1 , a 2 , a 3 , … a_1, a_2, a_3, \dots a 1 , a 2 , a 3 , … AP if the difference between consecutive terms is constant:
a n + 1 − a n = d (same for all n ) a_{n+1} - a_n = d \quad \text{(same for all } n\text{)} a n + 1 − a n = d (same for all n )
First term = a = a = a a 1 a_1 a 1
Common difference = d = a 2 − a 1 = a 3 − a 2 = … = d = a_2 - a_1 = a_3 - a_2 = \dots = d = a 2 − a 1 = a 3 − a 2 = …
Example: 2 , 5 , 8 , 11 , … 2, 5, 8, 11, \dots 2 , 5 , 8 , 11 , … a = 2 a=2 a = 2 d = 3 d=3 d = 3
WHY define it by the difference? Because "constant difference" is the only rule an AP obeys. Once you know it, every term is forced.
Intuition WHY there's a formulaTo reach term n n n a a a d d d 1 1 1 n n n ( n − 1 ) (n-1) ( n − 1 ) not n n n , because the first term costs zero jumps.
HOW — build it up:
a 1 = a ( 0 steps ) a 2 = a + d ( 1 step ) a 3 = a + 2 d ( 2 steps ) a 4 = a + 3 d ( 3 steps ) ⋮ a n = a + ( n − 1 ) d ( n − 1 steps ) \begin{aligned}
a_1 &= a &&(0 \text{ steps})\\
a_2 &= a + d &&(1 \text{ step})\\
a_3 &= a + 2d &&(2 \text{ steps})\\
a_4 &= a + 3d &&(3 \text{ steps})\\
&\;\;\vdots\\
a_n &= a + (n-1)d &&(n-1 \text{ steps})
\end{aligned} 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 worksWrite the sum forwards, then backwards underneath. Each vertical pair adds to the same value (first + last), because as one number climbs by d d d d d d n n n
HOW — the derivation:
Write the sum, then reverse it:
S n = a + ( a + d ) + ( a + 2 d ) + ⋯ + ( a + ( n − 1 ) d ) S n = ( a + ( n − 1 ) d ) + ⋯ + ( a + 2 d ) + ( a + d ) + a \begin{aligned}
S_n &= a + (a+d) + (a+2d) + \dots + \big(a+(n-1)d\big)\\
S_n &= \big(a+(n-1)d\big) + \dots + (a+2d) + (a+d) + a
\end{aligned} S n S n = a + ( a + d ) + ( a + 2 d ) + ⋯ + ( a + ( n − 1 ) d ) = ( a + ( n − 1 ) d ) + ⋯ + ( a + 2 d ) + ( a + d ) + a Add the two lines column by column . Every column gives 2 a + ( n − 1 ) d = a 1 + a n 2a + (n-1)d = a_1 + a_n 2 a + ( n − 1 ) d = a 1 + a n
2 S n = ( 2 a + ( n − 1 ) d ) + ⋯ + ( 2 a + ( n − 1 ) d ) ⏟ n equal terms = n ( 2 a + ( n − 1 ) d ) 2S_n = \underbrace{\big(2a+(n-1)d\big) + \dots + \big(2a+(n-1)d\big)}_{n \text{ equal terms}} = n\big(2a+(n-1)d\big) 2 S n = n equal terms ( 2 a + ( n − 1 ) d ) + ⋯ + ( 2 a + ( n − 1 ) d ) = n ( 2 a + ( n − 1 ) d ) Divide by 2:
WHY the second form is beautiful: the sum = (number of terms) × (average term). For an AP the average is just first + last 2 \frac{\text{first}+\text{last}}{2} 2 first + last
Worked example 1 — Find the 20th term of3 , 7 , 11 , … 3,7,11,\dots 3 , 7 , 11 , …
a = 3 a=3 a = 3 d = 7 − 3 = 4 d=7-3=4 d = 7 − 3 = 4 Why this step? d d d a 20 = a + ( 20 − 1 ) d = 3 + 19 ⋅ 4 a_{20} = a + (20-1)d = 3 + 19\cdot4 a 20 = a + ( 20 − 1 ) d = 3 + 19 ⋅ 4 Why 19 19 19 We jump n − 1 = 19 n-1=19 n − 1 = 19 = 3 + 76 = 79 = 3 + 76 = \mathbf{79} = 3 + 76 = 79
Worked example 2 — Sum of first 100 natural numbers
Here a = 1 a=1 a = 1 d = 1 d=1 d = 1 n = 100 n=100 n = 100 Why AP? Difference is constant = 1 =1 = 1
Last term a 100 = 1 + ( 99 ) ( 1 ) = 100 a_{100}=1+(99)(1)=100 a 100 = 1 + ( 99 ) ( 1 ) = 100
S 100 = 100 2 ( a + a 100 ) = 50 × 101 = 5050 S_{100}=\frac{100}{2}(a+a_{100})=50\times101=\mathbf{5050} S 100 = 2 100 ( a + a 100 ) = 50 × 101 = 5050 Why this form? We know first & last, so average form is fastest.
Worked example 4 — How many terms of5 , 8 , 11 , … 5,8,11,\dots 5 , 8 , 11 , … 84 84 84
a = 5 a=5 a = 5 d = 3 d=3 d = 3 S n = 84 S_n=84 S n = 84 84 = n 2 [ 2 ( 5 ) + ( n − 1 ) 3 ] = n 2 ( 3 n + 7 ) 84 = \frac{n}{2}[2(5)+(n-1)3] = \frac{n}{2}(3n+7) 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 = 0 168 = 3n^2+7n \Rightarrow 3n^2+7n-168=0 168 = 3 n 2 + 7 n ⇒ 3 n 2 + 7 n − 168 = 0 n = − 7 ± 49 + 2016 6 = − 7 ± 45 6 n=\frac{-7\pm\sqrt{49+2016}}{6}=\frac{-7\pm 45}{6} n = 6 − 7 ± 49 + 2016 = 6 − 7 ± 45 Why reject the negative root? n n n n = 38 6 n=\frac{38}{6} n = 6 38 n = − 52 6 n=\frac{-52}{6} n = 6 − 52 2065 ≈ 45.4 \sqrt{2065}\approx45.4 2065 ≈ 45.4 n ≈ 6.4 n\approx6.4 n ≈ 6.4 no exact n n n gives 84 with these values. (Lesson: always check the discriminant gives a whole number.)
n n n ( n − 1 ) (n-1) ( n − 1 ) Why it feels right: "20th term, 20 of something" — feels like multiply by 20.
The fix: The first term is free (0 jumps). Only n − 1 n-1 n − 1 d d d a + ( 1 − 1 ) d = a a+(1-1)d=a a + ( 1 − 1 ) d = a
d = a 2 / a 1 d = a_2/a_1 d = a 2 / a 1 Why it feels right: Confusing AP with GP, where you divide for the ratio.
The fix: AP = add a constant ⇒ d = a 2 − a 1 d=a_2-a_1 d = a 2 − a 1 multiply ⇒ ratio.
Common mistake Forgetting to checkn n n Why it feels right: You solved the quadratic, got a number — done!
The fix: n n n
Recall Master these three
a n = a + ( n − 1 ) d a_n = a+(n-1)d a n = a + ( n − 1 ) d S n = n 2 [ 2 a + ( n − 1 ) d ] = n 2 ( a + a n ) S_n = \frac{n}{2}[2a+(n-1)d] = \frac{n}{2}(a+a_n) S n = 2 n [ 2 a + ( n − 1 ) d ] = 2 n ( a + a n ) a n = S n − S n − 1 a_n = S_n - S_{n-1} a n = S n − S n − 1 a n a_n a n n n n
Recall Feynman: explain to a 12-year-oldImagine climbing stairs where every step is the same height . Your starting spot is a a a d d d n n n ( n − 1 ) (n-1) ( n − 1 ) = a + ( n − 1 ) d = a + (n-1)d = a + ( n − 1 ) d all the stair heights: line them up shortest to tallest, then place the same stairs tallest to shortest right beside them. Each pair now has the same total height. There are n n n = ( first + last ) = (\text{first}+\text{last}) = ( first + last ) = n ( first + last ) = n(\text{first}+\text{last}) = n ( first + last )
"nth term: subtract ONE before you step" → ( n − 1 ) d (n-1)d ( n − 1 ) d Sum = "How many × Average" → S n = n × first + last 2 S_n = n \times \frac{\text{first}+\text{last}}{2} S n = n × 2 first + last
Geometric Progression (GP) — multiply-by-ratio cousin of the add-by-d d d Arithmetic Mean — middle term of 3-term AP = a + c 2 = \frac{a+c}{2} = 2 a + c Sigma Notation — S n = ∑ k = 1 n a k S_n = \sum_{k=1}^{n} a_k S n = ∑ k = 1 n a k Quadratic Equations — used to solve "how many terms" problems.Gauss Summation Trick — the pairing idea behind S n S_n S n Linear Functions — a n a_n a n n n n
What defines an AP? The difference between consecutive terms is constant,
a n + 1 − a n = d a_{n+1}-a_n=d a n + 1 − a n = d .
Formula for the nth term of an AP? a n = a + ( n − 1 ) d a_n = a + (n-1)d a n = a + ( n − 1 ) d .
Why ( n − 1 ) d (n-1)d ( n − 1 ) d n d nd n d The first term needs 0 jumps; reaching term
n n n takes only
n − 1 n-1 n − 1 jumps of size
d d d .
Sum of n terms (standard form)? S n = n 2 [ 2 a + ( n − 1 ) d ] S_n = \frac{n}{2}[2a+(n-1)d] S n = 2 n [ 2 a + ( n − 1 ) d ] .
Sum of n terms (average/first-last form)? S n = n 2 ( a + a n ) S_n = \frac{n}{2}(a+a_n) S n = 2 n ( a + a n ) , i.e. terms × average of first and last.
How do you get a n a_n a n a n = S n − S n − 1 a_n = S_n - S_{n-1} a n = S n − S n − 1 .
If a n a_n a n n n n The sequence is an AP (its common difference is the slope).
If S n S_n S n n n n An AP.
Common difference of 2 , 5 , 8 , 11 2,5,8,11 2 , 5 , 8 , 11 d = 3 d = 3 d = 3 (subtract consecutive terms).
Trick used to derive S n S_n S n Write sum forwards + backwards and add; each column equals
a + a n a+a_n a + a n .
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
Intuition Hinglish mein samjhoDekho, Arithmetic Progression (AP) ka matlab bilkul simple hai: ek sequence jisme har baar tum same amount add karte ho. Wo fixed jump ko hum common difference d d d kehte hain. Jaise 2 , 5 , 8 , 11 2,5,8,11 2 , 5 , 8 , 11 + 3 +3 + 3 a a a d d d
nth term nikaalne ke liye socho: term 1 par pahunchne ke liye koi jump nahi chahiye (tum already wahi ho). Term n n n ( n − 1 ) (n-1) ( n − 1 ) a n = a + ( n − 1 ) d a_n = a+(n-1)d a n = a + ( n − 1 ) d n n n n − 1 n-1 n − 1 n = 1 n=1 n = 1 a a a
Sum ka jugaad Gauss wala hai. Sum ko ek baar seedha likho, ek baar ulta niche likho, aur column-wise add karo. Har column ka total same aata hai: a + a n a + a_n a + a n n n n 2 S n = n ( a + a n ) 2S_n = n(a+a_n) 2 S n = n ( a + a n ) S n = n 2 ( a + a n ) S_n = \frac{n}{2}(a+a_n) S n = 2 n ( a + a n ) Sum = kitne terms × average term .
Ye chapter isliye important hai kyunki EMIs, interest, seating rows, natural numbers ka sum (1 + 2 + ⋯ + 100 = 5050 1+2+\dots+100=5050 1 + 2 + ⋯ + 100 = 5050 a n a_n a n n n n straight line ho (linear), ya S n S_n S n quadratic ho bina constant ke, to sequence pakka AP hai. Bas teen formule ratt lo, baaki sab derive ho jaayega.
