Question

Find the sum of the first 40 positive divisible by 6

Class 10

Full explained

Note : No spam answer 5 answer will deleted​

Answer 1

Answer:

We know that the first 40 positive integers divisible by 6 are 6,12,18,....

This is an AP with a = 6 and d = 6.

S40= 20[2(6) +(40-1)6] =20[12+234] =4920.

Step-by-step explanation:

Hope you will understand

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\:the\;Question\;:-}}}$

Here the Concept of Arithmetic Progression has been used. We see when we arrange first 40 terms that are divisible by 6, the resultant sequence is AP. From that first, we can find the first term and common difference. Then, we can apply those in formula to find the answer.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{Common\;Difference,\;d\;=\;\bf{a_{2}\;-\;a_{1}\;=\;a_{3}\;-\;a_{2}}}}$

$\\\;\boxed{\sf{S_{n}\;=\;\bf{\dfrac{n}{2}[2a\;+\;(n\;-\;1)d]}}}$

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★ Solution :-

Given,

» First number to be divisible by 6 = 6

» Second number to be divisible by 6 = 12

» Third number to be divisible by 6 = 18

This series continues until we find the 40th number divisible by 6 .

Then the required A.P. will be,

$\\\;\;\sf{\mapsto\;\;A.P.\;:\;\pink{6,\;12,\;18,\;24,\;30,\;......,\;n}}$

Here, n = 40

Since, we need to find the sum of first 40 terms.

Here,

• First Term = a = 6
• Second Term = a₂ = 12
• Third Term = a₃ = 18

Now we need to find the Common Difference of this A.P.

$\\\;\;\sf{:\rightarrow\;\;Common\;Difference,\;d\;=\;\bf{a_{2}\;-\;a_{1}\;=\;a_{3}\;-\;a_{2}}}$

$\\\;\;\sf{:\rightarrow\;\;Common\;Difference,\;d\;=\;\bf{12\;-\;6\;=\;18\;-\;12}}$

$\\\;\;\sf{:\rightarrow\;\;Common\;Difference,\;d\;=\;\bf{6\;=\;6}}$

$\\\;\;\bf{:\rightarrow\;\;Common\;Difference,\;d\;=\;\bf{6}}$

Here both the conditions satisfy.

Hence, common difference = 6

Now sum of first 40 terms, is given as ::

$\\\;\;\sf{:\Longrightarrow\;\;S_{n}\;=\;\bf{\dfrac{n}{2}[2a\;+\;(n\;-\;1)d]}}$

This is sum of first n terms.

Replacing n by 40, we get

$\\\;\;\sf{:\Longrightarrow\;\;S_{40}\;=\;\bf{\dfrac{40}{2}[2a\;+\;(40\;-\;1)d]}}$

$\\\;\;\sf{:\Longrightarrow\;\;S_{40}\;=\;\bf{\dfrac{40}{2}\;\times\;[2(6)\;+\;(40\;-\;1)(6)]}}$

$\\\;\;\sf{:\Longrightarrow\;\;S_{40}\;=\;\bf{20\;\times\;[12\;+\;(39)(6)]}}$

$\\\;\;\sf{:\Longrightarrow\;\;S_{40}\;=\;\bf{20\;\times\;[12\;+\;(39)(6)]}}$

$\\\;\;\sf{:\Longrightarrow\;\;S_{40}\;=\;\bf{20\;\times\;[12\;+\;234]}}$

$\\\;\;\sf{:\Longrightarrow\;\;S_{40}\;=\;\bf{20\;\times\;246}}$

$\\\;\;\bf{:\Longrightarrow\;\;S_{40}\;=\;\bf{\green{4920}}}$

Hence, the sum of first 40 terms divisible by 6 = 4920

$\\\;\underline{\boxed{\tt{Thus,\;\;the\;\;required\;\;sum\;=\;\bf{\red{4920}}}}}$

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★ More formulas to know ::

$\\\;\sf{\leadsto\;\;\purple{a_{n}\;=\;a\;+\;(n\;-\;1)d}}$

Here aₙ = the nₜₕ term.

$\\\;\sf{\leadsto\;\;\purple{l\;=\;a\;+\;(n\;-\;1)(-d)}}$

Here l = the last term.

$\\\;\sf{\leadsto\;\;\purple{s_{n}\;=\;\dfrac{n(n\;+\;1)}{2}}}$

Here sₙ = sum of first n positive integers.