Question

1+5+9+13+17+................…+X=190.find X ?

Answer 1

The requried value of x= 145

Answer 2

Hi ,

1 + 5 + 9 + ..... , + X = 190

1, 5 , 9 , .... are in A.P

First term = a = 1

Common difference = a2 - a1

d = 5 - 1 = 4

Let the sum of 'n ' terms = 190

S = 190

n/2 [ 2a + ( n - 1 ) d ] = 190

n/2 [ 2 × 1 + ( n - 1 ) 4 ] = 190

2n/2[ 1 + 2n - 2 ] = 190

n ( 2n - 1 ) - 190= 0

2n² - n - 190 = 0

2n² - 20n + 19n - 190 = 0

2n( n - 10 ) + 19 ( n - 10 ) = 0

( n - 10 ) ( 2n + 19 ) = 0

Therefore ,

n - 10 = 0 or 2n + 19 = 0

n= 10 , n = -19/2

But n should not be fraction.

n = 10

Now ,

n th term = X

a + ( n - 1 ) d = X

1 + ( 10 - 1 ) 4 = X

1 + 36 = X

Therefore ,

X = 37

I hope this helps you.

:)