Question

The value of 1+3+5+7+…………………...+ 49 = *

Answer 1

Step-by-step explanation:

1,3,5,7,9......49 are in ap

a=1, d=5-3=3-1=2

an=a+(n-1)d

49=1+(n-1)2

49=1+2n-2

49=2n-1

n=(49+1)/2

n=25

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

=[(25/2){(2)(1)+(25-1)(2)}]

=[(25/2){2+48}]

=[(25)(25)]

=625

Answer 2

As we know that the first term of Arithmetic Progression is given as

1st Term of (A.P)⇒ $K=x+(z-1)y$

Here given, $K=49$  & $x=1 , y=2 , z=?$ ($z$ is the total number of term)

Now putting given value in Formula

⇒$K=x+(z-1)y$

⇒$49=1+(z-1)2$

⇒$49-1=(z-1)2\\48=(z-1)2\\z-1=\frac{48}{2} \\z-1=24\\z=25$

Hence the total number of terms in A.P is $z=25$

So to find Sum of Arithmetic Progression,

$S=\frac{z(x+1)}{2}$

Now putting values in the above equation,

$S=\frac{25(1+49}{2} \\S=\frac{25(50)}{2}\\S=25*25\\S=625$

Hence the Sum of Arithmetic Progression $S=625$