Math, asked by abhinav200410, 1 year ago

find the sum of even positive integers between 1 and 200​

Answers

Answered by student0135
1

Positive integers between 1 and 200 :-

2+4+6+8+10+12+14+16+18+20+22+24+26+28+30+32+34+36+38+40+42+44+46+48+50+52+54+56+58+60+72+74+76+78+80+82+84+86+88+90+92+94+96+98+100+102+104+106+108+110+112+114+116+118+120+122+124+126+128+130+132+134-136+138+140+142+144+146+148+150+152+154+156+158+160+163+164+166+168+170+172+174+176+178+180+182+184+186+188+190+192+194+196+198 = 9900

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Answered by varadad25
0

Answer:

The sum of the even positive integers between 1 and 200 is 9900.

Step-by-step-explanation:

The even positive integers between 1 and 200 are 2, 4, 6, 8,...., 196, 198.

Here,

  • t₁ = 2
  • t₂ = 4
  • t₃ = 6

t₂ - t₁ = 4 - 2 = 2

2t₃ - t₂ = 6 - 4 = 2

As the difference between two consecutive terms is constant, the sequence is an Arithmetic Progression ( AP ).

  • a = 2
  • d = 2
  • tₙ = 198

We have to find the sum of this AP.

We know that,

tₙ = a + ( n - 1 ) * d

⇒ 198 = 2 + ( n - 1 ) * 2

⇒ 198 - 2 = 2n - 2

⇒ 196 = 2n - 2

⇒ 2n = 196 + 2

⇒ 2n = 198

⇒ n = 198 / 2

n = 99

Now,

Sₙ = ( n / 2 ) [ 2a + ( n - 1 ) * d ]

⇒ Sₙ = ( 99 / 2 ) [ 2 * 2 + ( 99 - 1 ) * 2 ]

⇒ Sₙ = ( 99 / 2 ) 2 ( 2 + 98 )

⇒ Sₙ = 99 * 2 ÷ 2 * 100

⇒ Sₙ = 99 * 1 * 100

⇒ Sₙ = 99 * 100

Sₙ = 9900

The sum of the even positive integers between 1 and 200 is 9900.

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