Question

find the sum of last 14 terms multiple of 6 which are 4 digits​

Answer 1

Given:

• first term, a = 9996 ( Since it is the last 4-digit number that's divisible by 6 )
• common difference, d = -6 ( Because we are asked for the multiples of 6 )
• number of terms, n = 14

14th term from last = $l + (n-1)d$

$\Rightarrow \qquad 9996 + 13 \cdot -6 \\ \qquad \quad \; \boxed{l_{14} = 9918}$

We know that,

$\qquad \quad \; \boxed{S_{n} = \frac{n}{2}[ \: a + l \: ] } \\ \\ \qquad \quad \; S_{14} = \frac{14}{2}[ 9918 + 9996 \: ] \\ \Rightarrow \qquad 7 \times 19914 \\ \qquad \quad \; \boxed{S_{14} = 139398}$

$\huge{\mathtt{ANSWER \: : \: 139398 }}$

Answer 2

Answer:

139398

Step-by-step explanation:

last 4 digits number divisible by 6 is 9996

therefore a = 9996

difference between the number is 6

the AP will be

9996, 9990, 9984 .......

d = -6

Sn = n/2 (2a + (n-1)d)

$s14 = \frac{14}{2} (2(9996) + (14 - 1) - 6) \\ = 7(19992 - 78) \\ = 7 \times 19914 \\ = 139398$

hope you get your answer