Question

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answer according to class 10 .
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Answer 1

Step-by-step explanation:

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Answer 2

Question:-

Find sum of all 3digit no.s divisible by 11?

Step-by-step explanation:

Solution:-

At first divide 100 by 11

100=11×9+1

if we add 10 then we will get the 1st no. divisible by 11

i.e, 110

Now Divide 999 by 11

999=11×90+9

If we remove 9 then we will get the last no. divisible by 11

i.e,990

A.P↓

110,121,132,143.....990

a=110

d=a3-a2=132-121=11

°•°Let an=990

an=a+(n-1)d

990=110+(n-1)×11

990-110=(n-1)×11

880/11=(n-1)

80=n-1

n=80+1

n=81

Now we know there are 81 terms divisible by 11

so, S81=81/2[2a+(n-1)d]

=>81/2[2×110+(81-1)×11]

81/2[220+880]

81/2×1100

81×550

S21=44550

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