Question

Find the sum of first 40 positive integers divisible by 6​

Answer 1

GIVEN :–

• 40 positive integers which is divided by 6.

• 6 , 12 , 18 , .......... 40 terms.

TO FIND :–

• Sum of given terms = ?

SOLUTION :–

• We know that if a is first term and d is Common difference then , Sum of A.P. series –

$\\ \longrightarrow \: \large{ \boxed { \bold{ S_{n} = \dfrac{n}{2} \left \{2a + (n - 1)d \right \}}}}$

▪︎Here –

$\\ \: \: \: \: \:{ \huge{.}} \: \: \: { \bold{a = 6}}$

$\\ \: \: \: \: \:{ \huge{.}} \: \: \: { \bold{d = 6}}$

$\\ \: \: \: \: \:{ \huge{.}} \: \: \: { \bold{n=40}}$

• So that –

$\\ \implies \: { \bold{ S_{40} = \dfrac{40}{2} \{2(6) + (40 - 1)(6) \}}}$

$\\ \implies \: { \bold{ S_{40} = 20 \{12 + 39 \times 6 \}}}$

$\\ \implies \: { \bold{ S_{40} = 20 \{12 + 234 \}}}$

$\\ \implies \: { \bold{ S_{40} = 20 \{246 \}}}$

$\\ \implies \large{ \boxed { \bold{ S_{40} = 4,920}}}$

Answer 2

Step-by-step explanation:

${\red{\underline{\underline{\bold{Question:-}}}}}$

Find the sum of first 40 positive integers divisible by 6.

${\blue{\underline{\underline{\bold{Concept\:used\:here}}}}}$

• Sum of first n terms of an AP ( Arithmetic Progression)

${\green{\underline{\underline{\bold{Solution:-}}}}}$

Sum of first 40 integers divisible by 6

Given A.P:-

6,12,18,24..............40terms

In the given series:-

• a = 6

• n = 40

• d = $a _{2} - a$
• = 12 - 6 = 6

Where a is the first term of the A.P , n is the last term of the A.P and d is the common difference .

${\orange{\underline{\underline{\bold{Formula\:used:-}}}}}$

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

Substituting the values:-

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

$\implies \: S_{40} = \frac{40}{2} (2 \times 6 + (40 - 1)6)$

$\implies \: S_{40} = 20(12 + (39)6)$

$\implies\: S_{40} = 20( 12 + 234)$

$\implies \: S_{40} = 20( 246)$

$\implies \: S_{40} = 4920$

✏️$\boxed{S_{40} = 4,920}$

$\tt\purple{Therefore,\: The \: sum\: of \: first \: 40\: positive\: integers\: divisible \: by \: 6 \: is 4,920}$