Question

Find the sum of first 12 terms of the multiples of 2 and 3 using the formula

Answer 1

Answer

Since the question has two cases we will find one by one.

Now,

(1)To find sum of first 12 multiples of 2 >a= 2, d =2,n=12

Now, we know

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

=>Sn=$\frac{12}{2}$ [2×(2)+11d]

=>Sn=6[(4)+11(2)]

=>Sn = 6(4 + 22)

=>Sn = 6 × 26

             =>Sn = 156.       ------(i)

Hence the sum of First 12 terms of multiple of 2 is 156

2)To find sum of first 12 multiples of 3 >a= 3, d =3,n=12

Now, we know=

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

=>Sn = $\frac{12}{2}$ [2(3)+11d]

=>Sn = 6[(6) + 11×(3)]

=>Sn= 6(6 + 33)

=>Sn = 6 × 39

               =>Sn = 234.    -------(ii)

Hence the sum of First 12 terms of multiple of 3 is 234.

Hence,from (i) and (ii)=>

The sum of  first 12 terms of the multiples of 2 and 3 is 234+156 = 390