Question

11. What is the sum of all numbers between 250 and 550, which are divisible by 19.

(a) 6635
(b) 6640
(C) 6000
(d) 6695

::UPSC CAPF QUESTION​

Answer 1

Answer: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.

Step-by-step explanation: MARK ME AS THE BRAINLIEST PLEASE

Answer 2

The correct answer is 5985.

Given: Range of numbers 250 to 550.

To Find: Sum of all numbers between this range divisible by 19.

Solution:

So, 19*14 = 266 which is first number in this range.

19*14 = 266

19*15 = 285

19*16 = 304

19*17 = 323

19*18 = 342

19*19 = 361  upto 19*28 = 532

So the terms of series are,

266, 285, 304, 323, .........532

This is an AP. With common difference = 19.

First term = 266

Let total terms = n

$a_{n}=a+(n-1)d$

532 = 266 + (n-1)19

266 = (n-1)19

14 = n-1

n = 15

Sum of all numbers = $\frac{n}{2}(2a+(n-1)d)$

= $\frac{15}{2}(2*266+(15-1)19)$

= $\frac{15}{2}(532+(14)19)$

= $\frac{15}{2}(532+266)$

= $\frac{15}{2}(798)$

= 5985

Hence, the sum of all numbers between 250 and 550, which are divisible by 19 is 5985.

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