Find the sum of three-digit multipleis natural number , which is divis

1.	Find the sum of three-digit multipleis natural number , which is divisible by 11
Answer» Sorry there was some mistaken990=110+(n-1)11880=11n-1111n=891 so n=81Sn= 81/2( 2110+8011)=81/2 1100=44550 110,121,..........990Than A=110D=121-110=11An=990An =a+(n-1)d990=110+(n-1)11990=110+11n-11990=99+11nOr 11n=990-99n=891/11=81 Here first term a= 110, d=11 and last term =990suppose no. of terms =nthen 990= 110 +(n-1) x 11 (nth term of AP)\xa0n-1=8 so n=9sum of all the numbers= 9/2x( 2x110 + 8x11)=9/2 x ( 220+88) = 1386\xa0

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