Question

find the sum of all three digit numbers which are multiples of 11??

Answer 1

#RAM RAM ji ❤^_^

ĀNSWĒR ⏬⏬

The 3-digit numbers that are multiples of 11
are▶

110, 121,-------------990

so, a=110

common difference (d)=
a2- a1

=110-121

=11

an=990

so,

The number of terms is given by the formula for Arithmetic Progression is

 an=a+(n−1)d
where
an =990 ,a=110, d=11

so, put all the values in formula

i.e
an=a+(n-1)d

990=110+(n-1)11

990=110+11n-11

990=99+11n

990-90=11n

891=11n

n=891/11

n=81

Now,

Sn=n/2(a+an)

81/2(110+990)

81/2(1100)

=44550

THANKS ✌☺

#NAVI ❤ HARYANVI ♠

Answer 2

Answer:

44550

Step-by-step explanation:

All the three digit numbers which are multiple of 11 are 110,121,132,..990.

∴ This is an AP with:

⇒ First term = 110.

⇒ Common difference = 11

⇒ Last term = a(n) = l = 990.

nth term of an AP:

990 = a + (n - 1) * d

990 = 110 + (n - 1) * 11

990 = 110 + 11n - 11

990 = 99 + 11n

891 = 11n

n = 81.

Sum of n terms of an AP:

= n/2[a + l]

= 81/2[110 + 990]

= 81/2[1100]

= 81 * 550

= 44550.

Therefore, Sum is 44550.

Hope it helps!