Question

find the sum of all natural numbers between 250 and 1000 which are exacty divisible by 9

Answer 1

252+261+- - - - -+999
Tn=a+(n-1)d =84
Sn=n/2(2a+(n-1)d)
=42(1251)
=52542

Answer 2

Answer:

Sum of all Natural numbers between 250 and 1000 which are exactly divisible by 9 are 52542

Step-by-step explanation:

Natural numbers between 250 and 1000 which are exactly divisible by 9 are 252,261,...,999 is an A.P.

First term (a) = 252

last term (l) = 999

common difference (d)=9

l = 999

=> a+(n-1)d=999

=> 252+(n-1)d=999

Divide each term by 9, we get

=> 28 + n-1= 111

=> 27 + n = 111

=> n = 111 - 27

=> n = 84

Therefore,

Sum of 84 terms in A.P =Sn

$S_{n}= \frac{n}{2}[a+l]$

$\implies S_{84}=\frac{84}{2}[252+999]\\=42 \times 1251\\=52542$

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