Question

solve the problem....

Answer 1

Yo!
Your Answer !

Let three consecutive multiples of 11 be = x , x+ 11 , x+22

A/Q

x+x+11+x+22=363
3x+33=363
3x=363-33
3x=330
x=330/3
x=110

middle term = x + 11 => 121

So, Option B is correct

Answer 2

$\sf\underline{Step-by-step \: explanation}$

Let the common number be x

All the multiples of 11...

11 × 1 = 11

11 × 2 = 22

11 × 3 = 33

11 × 4 = 44

And so on...

$\sf\underline{According \: to \: the \: question...}$


x + x + 11 + x + 22 = 363

Why x + 11 and 22?

According to the multiples table, the consecutive multiples should be arranged in this way.

3x + 33 = 363

3x = 363 - 33

3x = 330

x = 110

x = 110

x + 11 = 110 + 11 = 121

x + 22 = 110 + 22 = 132


Hence the middle term is,

121 (Option B)