Find sum of all 2 digits natural which are multiple pf 11
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two digit no start from 10 and 11 is the first two digit no that is a multiple of 11 ; and 99 is the last two digit number , and also a multiple of 11
therefore we will take a=11. and an=99
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Answer:
495
Step-by-step explanation:
2 digit multiples of 11
AP: 11, 22, 33. . . . . . . .99
a= 11 d= 11
n= (l-a)/d + 1
= (99-11)/11 + 1
= 88/11 + 1
= 8 + 1
=9
sₓ = (n/2)(a+l)
s₁₁ = (9/2)/(11+99)
= (9/2)/(110)
= 9×55
=495
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