Question

find the sum of last 14 four digits multiple of 6​

Answer 1

SOLUTION:

Given:

• first term, a = 6 ( Because it is the first multiple of 6 )
• common difference, d = 6 ( Since we are asked for the multiples of 6 )
• number of terms, n = 14 ( Given in the question )

We know that,

$\qquad \quad \; \boxed{S_{n} = \frac{n}{2}[ \: 2a + (n-1)d\: ]} \\ \\ \Rightarrow \qquad S_{14} = \frac{14}{2}[\: 2\cdot 6 + (14-1)\cdot 6 \: ] \\ \Rightarrow \qquad S_{14} = 7(12 + 13\cdot 6) \\ \Rightarrow \qquad S_{14} = 7 \cdot 90 \\ \qquad \quad \; \boxed{S_{14} = 630}$

$\huge{Answer \: : \: 630}$

Answer 2

Answer:

Given:

first term, a = 6 ( Because it is the first multiple of 6 )

common difference, d = 6 ( Since we are asked for the multiples of 6 )

number of terms, n = 14 ( Given in the question )

We know that,

\begin{gathered}\qquad \quad \; \boxed{S_{n} = \frac{n}{2}[ \: 2a + (n-1)d\: ]} \\ \\ \Rightarrow \qquad S_{14} = \frac{14}{2}[\: 2\cdot 6 + (14-1)\cdot 6 \: ] \\ \Rightarrow \qquad S_{14} = 7(12 + 13\cdot 6) \\ \Rightarrow \qquad S_{14} = 7 \cdot 90 \\ \qquad \quad \; \boxed{S_{14} = 630}\end{gathered}

S

n

=

2

n

[2a+(n−1)d]

⇒S

14

=

2

14

[2⋅6+(14−1)⋅6]

⇒S

14

=7(12+13⋅6)

⇒S

14

=7⋅90

S

14

=630

\huge{Answer \: : \: 630}Answer:630