Question

find the sum of all two digit natural numbers which are multiple of 6​

Answer 1

Answer:

12,18,24 ,30........96

here t1=12 ,t2=18

d=t2-t1

18-12=6

TN=a+(n-1)d

96=12+(n-1)6

96-12=6n-6

84-6=6n

78=6n

78÷6=n

13=n

sn=n÷2[2a+(n-1)d]

sn=13÷2[2×12+(13-1)6]

13÷2[24+72]

13÷2×96

13×48

624

sum of two digits natural number divided by 6 is 624

Answer 2

To Find :

The sum of all two-digit natural numbers which are multiples of 6​

Solution:

12,18,24 ,30........96  are the two-digit natural numbers which are multiples of 6​

Here,$t_{1} =12,t_{2} = 18$

Difference ,$d = t_{2} - t_{1}$

                  d= 18 - 12

                   d=6

The formula is given by,

$t_{n}$=a+(n-1)d

96=12+(n-1)6

96-12=6n-6

84+6=6n

90=6n

n = $\frac{90}{6}$

n = 15

$S_{n}$=$\frac{n}{2}$[2a+(n-1)d]

$S_{15}$=$\frac{15}{2}$[2×12+(15-1)6]

$S_{15}$=$\frac{15}{2}$[24+(14)6]

$S_{15}$=$\frac{15}{2}$[24+84]

$S_{15}$=$\frac{15}{2}$[108]

$S_{15}$ = 15 ×54

$S_{15}$= 810

Therefore, the sum of all two digit natural numbers which are multiple of 6​ is 810