BITSAT Mathematics Question (2021) — Solution

Step-by-step solution

Sum of odd numbers between 1 and 1000 , which is divisible by 3=3+9+15+21+27+ +999= S (let) Let n be the number of terms in series and a is first term. l=a+(n-1) d, where l is last term and d is is common difference. array l 999=3+(n-1) 6 \\ n-1= 999-3 6 = 996 6 \\ n-1=166 \\ n=167 \\ S = n 2 [2 a+(n-1) d] \\ = 167 2 [2 3+(167-1) 6] \\ = 167 2 [1002]=167 501=83667 array