Page 1 :
Selection of Terms :, Taken 3 terms in A.P. :, (a-d), a, (a+d), Taken 4 terms in A.P. :, (a - 3d), (a- d), (a+ d), (a + 3d), Taken 5 terms in A.P. :, , (a - 2d), (a-d), a, (a+ d), (a+ 2d), , Sum of n Terms }———————, , n*Term )—pl T, = a+ (rel) d.| |, , S,= a, + a, +a, *...., , , , | a,>a,+d=5— 2” Term, | a, a,+2d= 8 > 3% Term, , | a,34,+(nl)d= 74 25" Term |, , Sum of 1° n terms of an A.P. > S,, , +a, + 4,., , S.= 5 [2a + (n-1)d] = ; [a, + 4]., , Arithmetic, Progressions, , *, , , , , PROPERTIES, , If any nth term of a sequence is a linear expression, inn e.g. a, = An+ B, then the given sequence is an A.P., . If a constant term is added to or subtracted from, each term of an A.P. then the resulting sequence, is also an A.P. with the same common difference., . If each term of agiven A.P. is multiplied or divided, by a non-zero constant K, then the resulting sequence, is also an A.P. with common difference Kd or, respectively. Where d is the common difference of, the given AP., . Ina finite AP. the sum of the terms equidistant from, , the beginning and end is always same and is equal, to the sum of Ist and last term., , RESULTS ), , i, , S,, = 25 [2(2) + (25-1)3] = 950, 2, , Sum of n natural nos., , eg. If 2,58......are in AP. then, a=2andd=3, , , , 2, e.g. Sum of first ie, 4 7 natural nos. => S=28, et (n—m + 1)" term from start, , m" term from end, , Find A.P. whose n*, term is given ?, , , , , , T,, eg. S,=n'+2n, ; ss eenei, , Find T, when S, is given =", , , , are 3 terms, , | of an AP. then:, j atc =2b., , Condition of anA.P. | ee, = If, a,b,c, , NCERT / X / Arithmatic progression
