Question

find the sum of all natural numbers between 250 and 1000 diviaible by 3​

Answer 1

Answer:

Step-by-step explanation:

The first number after 250 divisible by 3 is 252. and the number just before 1000 divisible by 3 is 999. So there are 333–84+1 = 250 numbers between 250 and 1000 divisible by 3. Their sum = (n/2)(a+l) = (250/2)(252+999) = 156375.

Answer 2

We know that,

☯ The number between 250 and 1000 which are divisible by 3 are 250, 255, 258,......,999.

$\sf Here \begin{cases} & \sf{First\;term,\;a = 252 } \\ & \sf{Common\; difference,\;d = 3} \\ & \sf{Last\;term,\;l = 999} \end{cases}$

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$\star{\boxed{\sf{\purple{a_n = a + (n - 1)d}}}}$

$:\implies\sf a + (n - 1)d = 999$

$:\implies\sf 252 + (n - 1) \times 3 = 999$

$:\implies\sf 252 + 3n - 3 = 999$

$:\implies\sf 3n + 249 = 999$

$:\implies\sf 3n = 999 - 249$

$:\implies\sf 3n = 750$

$:\implies\sf n = \cancel{ \dfrac{750}{3}}$

$:\implies{\boxed{\frak{\pink{n = 250}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Number\;of\;terms,\;n\;is\;\bf{250}.}}}$

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☯ Sum of natural numbers between 250 and 1000 is,

$\star{\boxed{\sf{\purple{S_n = \dfrac{n}{2} \bigg( a + l \bigg)}}}}$

$:\implies\sf \dfrac{250}{2} \bigg( 252 + 999 \bigg)$

$:\implies\sf 125(1251)$

$:\implies{\boxed{\frak{\pink{156375}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\; required\;sum\;of\; natural\;numbers\;between\;250\;and\;1000\;is\; \bf{156275}.}}}$

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$\qquad\boxed{\underline{\underline{\bigstar \: \bf\:Formula\:Related\:to\:AP\:\bigstar}}}$

⠀⠀⠀⠀⠀⠀⠀

$\sf (i)\;The\; n^{th}\;term\;of\;an\;AP\; = \; \red{a_n + (n - 1)d}$

⠀⠀⠀⠀⠀⠀⠀

$\sf (ii)\;Sum\;of\;n\;term\;of\;an\;AP\; = \; \purple{S_n = \dfrac{n}{2} \bigg\lgroup\sf 2a + (n - 1)d \bigg\rgroup}$

⠀⠀⠀⠀⠀⠀⠀

$\sf (iii)\;Sum\;of\;all\;terms\;of\;AP\;having\;last\:term\;as\;'l'\; = \; \pink{ \dfrac{n}{2}(a + l)}$