Question

Find the sum of integers from 11 to 1000 which are divisible by 3

Answer 1

AP: 12,15,18........ 999
a=12, d=3,an=999, n=?,
an= a+(n-1)d
999=12+(n-1)3
987=(n-1)3
329=n-1
n=330
Now using the sum formula,
n=330,a=12,an=999
Sn=n/2[a+an]
=330/2(12+999)
=165*1011
=166815
This is the answer!!

Answer 2

Solution :

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Given :
To Find the sum of the numbers divisible by 3, between 11 to 1000,.

∴  a = 12,

∴ d = 3,.

$S_n = ?$
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To find $S_n = ?$ we need to find number of terms :

⇒ [tex]a_n = a + (n - 1)d [/tex]

⇒ $1000 = 12 + (n - 1)3$

⇒ $1000 - 12 = (n-1)3$

⇒ 988 = (n - 1) 3

⇒ $n - 1 = \frac{988}{3}$

⇒ $n - 1 = 329 \frac{1}{3}$

⇒ $n = 329 \frac{1}{3} + 1$

⇒ $n = 330 \frac{1}{3}$

⇒ n = 330 (as 1000 isn't divisible by 3⇒ $a_n = 999$)

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⇒ $S_n = \frac{n}{2} (a + l)$

⇒ $S_n = \frac{330}{2}(999 + 12)$

⇒ $S_n = 165(1011)$

⇒ $S_n = 166815$

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                                        Hope it Helps !!

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