Find the sum of all two digit positive integers which are divisible by 3 but not by 2
Answers
Answered by
32
15 is the first and 99 is the last two digit number which are divisible by 3 but not by 2.
So first term(a) = 15 and last term(l) = 99 and also common difference (d) = 6
Also, nth term= 99
or, a+(n-1)d = 99
or, 15+(n-1)6= 99
or, 15+6n-6 = 99
or, 6n = 90
so n = 15
Hence, Sum = 15/2{15+99}
or, sum = 15/2× 114
or,, sum = 855
Hence the answer is 855.
Hope it helped
plss mark as brainliest.
So first term(a) = 15 and last term(l) = 99 and also common difference (d) = 6
Also, nth term= 99
or, a+(n-1)d = 99
or, 15+(n-1)6= 99
or, 15+6n-6 = 99
or, 6n = 90
so n = 15
Hence, Sum = 15/2{15+99}
or, sum = 15/2× 114
or,, sum = 855
Hence the answer is 855.
Hope it helped
plss mark as brainliest.
Anonymous:
Thanx uday bro
Answered by
11
It can be expressed as an AP:-
15,21,27.....99
Here a= 15 and d=6
So An=99= 15+ (n-1)6
or n= 15
So S15= 15/2 { 2×15+ 14× 6} = 855
15,21,27.....99
Here a= 15 and d=6
So An=99= 15+ (n-1)6
or n= 15
So S15= 15/2 { 2×15+ 14× 6} = 855
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