Page 1 :
Achunga Eg.(i) ah khian : first term (a) = 1, common difference (d) = 4- 1=7-4=3;, Eg (ii) ah khian : first term (a) = 18, common difference (d) = 16-18= 14-16 =-2, Eg (iii) ah khian : first term (a) =— 7, common difference (d) = — 3 — (-7) =-3 + 7=4., ** n' term of an A.P._[ Term engzahna nge anih ? tih leh Term engzatnge awm ? tih zawnna formula], Ta =a+(n— oa fn term from the aia, ** First term(T)) = a,Second term(T2)= a + d, Third term(T3)= a + 2d, Fourth term(T4)= a + 3d and so on., [*** (n-1) awmzia chu term zat aiin pakhat (1) zelin a tlem tihna ani mai.|, , , , , , Example: Find the 13" term of the A.P. 12,17,22.......... [ 12 a intanin, 5 zelin pung ta se, 13-na tur enge ? ], Solution: Here first term (a) = 12, Common difference (d) = 17-12 =5 , n= 13, , Wehave, T, =a+(n-—1)d, Ti3 = 12+(13-1)(5) d aia number hi bracket chhungah dah, , = 12+ 12(5) zel tur, (n—1) hi d in kan puntir tihna ani, a, minus aniha in buai loh nan, , , , =12+60, =72. Thus, 13" term is 72, , , , , , , , Example: Find the 9" term of the A.P. 15,12,9,6,,......... [ 15 a intanin, 3 zelin tlem ta se, 9 -na tur enge ? ], Solution: Here first term (a) = 15, Common difference (d) = 15-12 =-3,n=9, We have, T, =a+(n-1)d, Ty = 15+(9-1)(-3), =15+8(-3), =15-24=-9 [ Thus 9" term is —9 ], Example: How many terms are there in the A.P. 12,17,22........., 82.? [ 12 a intanin, 5 zelin pung ta se, 82 thleng, hian term engzatnge awm ang? emaw 82 hi term engzahna nge?], Solution: Here first term (a) = 12, Common difference (d) = 17-12=5, n' term (T,) = 82, We have, Ty =a+(n—-1)d, => 82 = 12+(n-1)(5), => 82=12+5n-5, 7, => 82-12 +5=5n =>75 =5n =>n= 3 TIS: Thus, there are 15 terms., Example: Which term of the A.P. 21,18,15, .... is— 81., Solution: Here first term (a) = 21, Common difference (d) = 18-21 =-3,, , n" term (T,) =- 81 Wehave, T, =a+(n—1)d { 21 a intanin, 3 zelin tlem ta, => -81=21+(n-1)(-3) se, -81 hi term engzah nge?], => -81=21-3n+3, =>-81-21-3=-3n, , , , =>-105 =-3n, -105 . . :, > n= =3 =35 Thus — 81 is its 35" term., Example: How many two digit numbers are divisible by 3? __[ Digit hnih nei number (i.e.10 — 99) pathumin a, Solution: Two digit numbers divisible by 3 are 12,15,18, ....99 sem theih eng zatnge awm?], Here first term (a) = 12, Common difference (d) = 15-12=3, n" term (T,) = 99, Wehave, TT, =at+(n-1)d => 99 = 12+(n-1)(3), => 99=12+3n-3 => 99-12 +3=3n, => 90 =3n = n= = =30 Thus, there are 30 numbers., Example: The 7" term of an AP is 32 and its 13" term is 62. Find the AP. [ Hei hi Linear equations in, Solution : Let ‘a’ be the first term and ‘d’ be the common difference. two variables kha ania, ‘a’, 7 term = 32 and 13" term = 62 coefficient hi a inan sa, ie. at+(7-1)d =32 ie. at+(13—1)d=62 avangin kan paih nghal mai, or at+6d=32.............@) or at+12d=62... i, , , , ang, Subtracting (i) from (ii) we get 6d=30 => d=30/6 => d=5, Putting d=5 in equation (i) we get a+6(5)=32 => a+30=32 =>a=32-30 =>a=2, hus the AP is: 257,02.0 7. s.ce003, , Example: Find the 12" term from the end of the A.P. 4,9,14,........, 254 [ A tawp lam atangin term 12 na enge?], Solution: Here first term (a) = 4, Common difference (d) = 9 -— 4=5 , n= 12, last term (J) = 254, Wehave, T, =/-(n-1)d, Ti = 254-(12- 1)(5) =254-11(5) =254-55=199 [Thus 12" term from the end is 199]