Question

FIND THE SUM OF THE FIRST 25 EVEN NATURAL NUMBER

Answer 1

Answer:

Step-by-step explanation:

First 25 even natural no.s

S={2,4,6..........48,50}

tn=a+(n-1)d

50=2+(n-1)2

50=2n

n=25

Sn=n/2[2a+(n-1)d]

S25=25/2[2×2+24×2]

S25=650

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Answer 2

The sum of first 25 even natural numbers are 650.

Given:

First 25 even natural numbers are 2, 4, 6 , 8 ..........48,50

To find:

The sum of first 25 even natural numbers

Formula used:

Sum of nth terms  ,$S_ {n}$ = $\frac{n}{2} [a + l]$

Here,  a = first term , n = number of terms, l = last term

Solution:

Step 1: Substitute the values of a, n and l in the above formula from the first 25 even natural numbers and solve:

First 25 even natural numbers are 2,4,6 , 8 ..........48,50

Here,  first term , a = 2, number of terms, n = 25, last term, l = 50

Sum of nth terms , $S_ {n}$ = $\frac{n}{2} [a + l]$

⇒ $S_ {25}$ = $\frac{25}{2} (2 + 50)$  

⇒ $S_ {25}$ = $\frac{25}{2} (52)$

⇒ $S_ {25}$ = 25 × 26

⇒ $S_ {25}$ = 650

Hence, the sum of first 25 even natural numbers are 650.

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