Question

find the sum of all natural number between 250 and 1000 which are divisible by 3

Answer 1

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999 
This is an A.P. whose first term a = 252, Common difference, d = 3 and Last term l=999
We Know that 
l=a+(n-1)d
999=252+(n-1)3
999 - 252 = 3(n - 1)
n=250
We know that 
S=n/2(a+l)
=250/2(252+999)
=125*1251
=156375
Hence, the required sum is 156375.(Ans)

Answer 2

⠀⠀⠀⠀⠀⠀✇ The Number between 250 and 1000 which are exactly divisible by 3 are 252, 255, 258,....,999.

Therefore, This is an AP with,

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

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$\underline{\bigstar\:\boldsymbol{According\:to\: Question\::}}$

$\qquad \quad:\implies\sf a_n = 999$

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

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

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

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

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

$\: \: \: \quad \qquad:\implies\sf 3n = 750$

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

$\quad \qquad:\implies{\underline{\boxed{\frak{\purple{n = 250}}}}}\;\bigstar$

$\therefore\;{\underline{\sf{There\;are\;250\; natural\;numbers\;between\;250\;and\;1000.}}}$

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✇ Now, Finding sum of 250 terms of AP,

$\dag\;{\underline{\frak{We\;know\;that,}}}$

$\qquad\qquad \: \: \star{\boxed{\sf{\pink{S_n = \dfrac{n}{2} \bigg\lgroup\sf a + l \bigg\rgroup}}}}$

$\qquad:\implies\sf \dfrac{\cancel{ 250}}{ \cancel{2}} \bigg\lgroup\sf 252 + 999 \bigg\rgroup$

$\qquad\qquad:\implies\sf 125 \bigg\lgroup\sf 1251 \bigg\rgroup$

$\qquad\qquad \: \: \: :\implies{\underline{\boxed{\frak{\purple{156375}}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Sum\;of\;all\;natural\;numbers\;between\;250\;and\;1000\;is\; \bf{156375}.}}}$