What will be the sum of all the even numbers from 41 to 80?1. 20802. 6

1.	What will be the sum of all the even numbers from 41 to 80?1. 20802. 6403. 16404. 1220
Answer» Correct Answer - Option 4 : 1220 Given Numbers from 41 to 80 All number between 41 to 80 should be even number Formula used T_n = a + (n-1)d S_n = \(\frac{n}{2}\left\{ {2A + \left( {N - 1} \right)D} \right\}\) Where, T_n = last number in the series S_n = Sum of series of AP N = number of terms in series A = first number of the series D = difference of consecutive series Calculation: Even number between 41 to 80 ⇒ 42, 44, 46, 48, 50 ..............80 ⇒ This series is in AP Where, a = 42 ⇒ d = a₂ - a₁ _⇒44 - 42 = 2 T_n = a + (n-1)d ⇒ 80 = 42 + (n-1)2 ⇒ 38 = (n-1)2 ⇒ n = 20 ⇒ S_n= \(\frac{{20}}{2}\left\{ {2 × 42 + \left( {20 - 1} \right)2} \right\}\) ⇒ 20{42 + 19} ⇒ 20 × 61 ∴ 1220

What will be the sum of all the even numbers from 41 to 80?1. 20802. 6403. 16404. 1220

Answer» Correct Answer - Option 4 : 1220

Given

Numbers from 41 to 80

All number between 41 to 80 should be even number

Formula used

T_n = a + (n-1)d

S_n = \(\frac{n}{2}\left\{ {2A + \left( {N - 1} \right)D} \right\}\)

Where, T_n = last number in the series

S_n = Sum of series of AP

N = number of terms in series

A = first number of the series

D = difference of consecutive series

Calculation:

Even number between 41 to 80

⇒ 42, 44, 46, 48, 50 ..............80

⇒ This series is in AP

Where, a = 42

⇒ d = a₂ - a₁

_⇒44 - 42 = 2

T_n = a + (n-1)d

⇒ 80 = 42 + (n-1)2

⇒ 38 = (n-1)2

⇒ n = 20

⇒ S_n= \(\frac{{20}}{2}\left\{ {2 × 42 + \left( {20 - 1} \right)2} \right\}\)

⇒ 20{42 + 19}

⇒ 20 × 61

∴ 1220