source: ncert
topic: algebra
class_level: class_10
chapter: ch5
difficulty: intermediate
Class 10 | Ch5: Arithmetic Progressions ARITHMETIC PROGRESSION (AP): sequence where difference between consecutive terms is constant. Common difference d=a2-a1=a3-a2=... (can be positive, negative, or zero). General form: a, a+d, a+2d, a+3d, ... nth TERM FORMULA: a_n = a + (n-1)d where a=first term, d=common difference. SUM OF FIRST n TERMS: S_n = n/2 × [2a + (n-1)d] = n/2 × [a + l] where l=last term=a_n. FINDING NUMBER OF TERMS: use a_n=a+(n-1)d, solve for n. PROPERTIES: If a,b,c are in AP: b-a=c-b → 2b=a+c → b=(a+c)/2 (b is arithmetic mean). Sum of AP from both ends: a1+a_n=a2+a_(n-1)=... (equidistant terms have equal sum). THREE NUMBERS IN AP: take as a-d, a, a+d (sum=3a, simplifies problems). FOUR NUMBERS IN AP: take as a-3d, a-d, a+d, a+3d. SOLVED EXAMPLES: Example 1: Sum of first 40 positive integers divisible by 6. AP: 6,12,18,...,240. a=6, d=6, n=40. S_40=40/2×[2(6)+(40-1)(6)]=20×[12+234]=20×246=4920. Example 2: 7th term=34, 13th term=64. Find AP and S_20. a+6d=34 ...(i). a+12d=64 ...(ii). (ii)-(i): 6d=30 → d=5. From (i): a=34-30=4. AP: 4,9,14,19,... S_20=20/2×[2(4)+19(5)]=10×[8+95]=10×103=1030. Example 3: How many terms of AP 3,5,7,... give sum 120? S_n=n/2×[2(3)+(n-1)(2)]=n/2×[4n+2]=n(2n+1)... Wait: n/2[6+2n-2]=n/2[2n+4]=n(n+2)=120. n^2+2n-120=0. (n+12)(n-10)=0. n=10. COMMON MISTAKES: nth term formula: a_n=a+(n-1)d NOT a+(n)d. In sum formula S_n=n/2[2a+(n-1)d], the n inside brackets is (n-1) not n. d can be negative (decreasing AP) or zero (constant sequence).