Math, asked by rajat234015, 1 year ago

100 ते 200 पर्यंतच्या सर्व संख्यांची बेरीज किता ?
1) 15150
2) 1515
2
5 मध्ये किती मिळवावेत म्हणजे बेरीज 4 येईल ?
19
1) 7
2) 26
एका कोनाचा कोटिकोन त्याच्या पूरक कोनाच्या
1) 15°
2) 25°​

Answers

Answered by gopal8249
6

Step-by-step explanation:

(1)15150

(2)19

(3)25 answer for this questions

Answered by amitnrw
2

100 ते 200 पर्यंतच्या सर्व संख्यांची बेरीज   15150

Step-by-step explanation:

100 ते 200 पर्यंतच्या सर्व संख्यांची बेरीज किता

a = 100

d = 1

l = 200

l = a + (n-1)d

200 = 100 + (n-1)1

100 = n - 1

n = 101

S = (n/2)(a + l)

= (101/2)(100 + 200)

= 150 * 101

= 15150

5 मध्ये किती मिळवावेत   म्हणजे बेरीज 4 येईल

5 + x = 4

=> x = 4 - 5

=> x = -1

5 मध्ये  -1 मिळवावेत

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