Question

If 1+5+9+.......+x=780 what is the value of x

Answer 1

Sn=n(2a+(n-1)d)/2. [1]
an=a+(n-1)d. [2]
Sn=780
a=1
d=4
an=x
from [1]
780=n(2(1)+(n-1)4)/2=>n(2+4n-4)/2=>n(4n-2)/2
780= 2n^2-n
2n^2-n-780=0
780=40×39
2n^2-40n+39n-780=0
2n(n-20)+39(n-20)=0
(2n+39)(n-20)=0
n=20 or -39/2
n=20
from[2]
x=1+(20-1)4
x=1+19×4
x=1+76
x=77

Answer 2

Given,

1+5+9+      +x=780

To Find,

The value of x =?

Solution,

We can solve the question using the following steps:
Considering the given series, 1+5+9+     +x=780

The difference between the numbers,

5 - 1 = 4

9 - 5 = 4

Therefore, the given series is in Arithmetic Progression.

Now,

Sum of terms in AP = $\frac{n}{2} (2a + (n - 1)d)$   ------------ (1)

Here,  $d =$ common difference

          $n =$ the total number of terms

           $a =$ the first term

To find $n,$

$T_{n} = a + (n - 1)d$

$T_{last} = x = 1 + (n - 1)4$

        $x = 1 + 4n - 4$

         $x = 4n - 3$

         $x + 3 = 4n$

         $n = \frac{x + 3}{4}$

Now, substituting the given values in equation (1),

$780 = \frac{\frac{x + 3}{4} }{2} (2*1 + (\frac{x + 3}{4} - 1)4)$

$780 = \frac{x + 3}{8} (2 + (\frac{x - 1}{4} )4)$

$780 = \frac{x + 3}{8} (2 + x -1)$

$780 = \frac{x + 3}{8} (1 + x)$

$780 =\frac{x + x^{2} + 3 + 3x}{8}$

$6240 = x^{2} + 4x + 3$

$x^{2} + 4x - 6237 = 0$

Solving the above question, the roots are -81 and 77. Since, the value cannot be negative, x = 77.

Hence, the value of x is equal to 77.