find the sum of all the two digit number which are either multiples of 2or 3.
AlienRapper:
U mean we can chose either 2 or 3??
Or both??
Both are compulsory
Oh.. kk
I am writing the answer..
Ok check my answer whether it is wrong or right
ur answer is wrong man!
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1st 2 digit no which is a multiple of 2 = 10
i.e., a = 10
Last 2 digit no. divisible by 2 = 98
i.e., an = 98
Common difference, d = 2
an = a + (n - 1)d
98 = 10 + (n -1)2
98 - 10 = (n - 1)2
88 = 2n - 2
2n = 88 + 2
2n = 90
n = 90/2
n = 45
Therefore, Sn = n/2(a + an)
Sn = 45/2(10 + 98)
= 45/2 x 108
= 45 x 54
= 2430 is the answer.
Similarly, solve for multiples of 3.
Hope this helps...
Plz mark as brainliest if it helped...
Glad to help :)
i.e., a = 10
Last 2 digit no. divisible by 2 = 98
i.e., an = 98
Common difference, d = 2
an = a + (n - 1)d
98 = 10 + (n -1)2
98 - 10 = (n - 1)2
88 = 2n - 2
2n = 88 + 2
2n = 90
n = 90/2
n = 45
Therefore, Sn = n/2(a + an)
Sn = 45/2(10 + 98)
= 45/2 x 108
= 45 x 54
= 2430 is the answer.
Similarly, solve for multiples of 3.
Hope this helps...
Plz mark as brainliest if it helped...
Glad to help :)
THe answer now also is wrong!
He said we have to solve for both 2 and 3...
No... Ur wrong
Okkk I have No time to waste on uhhhh
Yes... No time to waste on proving my right answer is wrong and no time to prove that ur wrong answer is right...
Okay whatever u thinks I don't care
I know I am right and I Don't need to prove myself
U r 100%wrong bro...!
Mine is marked brainliest... :p
Kk iDc
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