Question

The sum of all integers between 200 and 800, which is divisible by 9 is (a)32769 (b)32796 (c)32976 (d)32967

Answer 1

Answer:

32,967‬

Step-by-step explanation:

200/9 = 22.222

=> 1st integer Divisible by 9 after 200

= 23 * 9  = 207

800/9 = 88.8888

=> Last integer before 800 divisible by 9

= 88 * 9 = 792

207 + 216 + ............................................+ 783 + 792

= 23*9  + 24*9  +..................................+ 87*9  + 88*9

= 9(23 + 24 +..........................................+ 87 + 88)

= 9 (66/2)(23 + 88)

= 9 * 33 * 111

= 32,967‬

Answer 2

The sum of all integers between 200 and 800, which is divisible by 9 is "(d) 32967".

Step-by-step explanation:

The sum of all integers between 200 and 800,which is divisible by 9 are:

207, 216, 225, ......., 792

The above sreies are in AP.

Here, first term (a) = 207, common difference(d) = 216 - 207 = 9 and

last term(l) = 792

Let number of term = n

∴ $a_{n} =a+(n-1)d$

⇒ $207+(n-1)9=792$

⇒ $(n-1)9=792-207=585$

⇒ $(n-1)=65$

⇒ n = 65 + 1 = 66

∴ The sum of all integers between 200 and 800, which is divisible by 9

= $\dfrac{n}{2} (a+l)$

$=\dfrac{66}{2} (207+792)$

$=33 (999)$

= 32967

Hence, the sum of all integers between 200 and 800, which is divisible by 9 is "(d) 32967".