Question

please answer this ​

Answer 1

3+6+9+12+...........+99.

in this series AP.

a=3,d=3,last number,l=99.

sum=n/2[a+l],

l=a+(n-1)d

99=3+(n-1)3

96/3=n-1

32=n-1

n=33,

sum=33/2[3+99]=33×51=1683.

Answer 2

here odd integers divisible by 3 between two to 100 are:

3,9,15,21,27..........99

therefore,

common difference= 6

also,

a= first term= 3

therefore, the numbers are in AP

now,.

tn= a+(n-1)×d

99= 3+ (n-1)×6

99= 3+ 6n-6

99= -3+ 6n

99+3= 6n

102= 6n

102/6= n

17= n

therefore, there are 17 odd integers divisible by 3 from 2 to 100

now,

according to the formula,

Sn= n/2×[2a+(n-1)×d]

= 17/2×[2×3+(3-1)×6]

= 17/2×[6+2×6]

= 17/2×18.

= 17×9

= 153

therefore, sum of odd integers dividible by 3 from 2 to 100= 153

hope this helps!!!✌☺☺❤