Question

The minimum no of terms of 1+3+5+7+.....that add up to a number exceeding 1357 is

Answer 1

a=1
d=2
An=1357
n=?
An=a+(n-1)d
1357=1+(n-1)2
1356=2n-2
1358=2n
n=679

Answer 2

Answer :

Given :

a ( first number ) = 1

d ( Difference ) = 2

Aⁿ ( Final term ) = 1357

n ( Total Numbers ) = ?

Solution :

Aⁿ = a + ( n - 1 ) d

1357 = 1 + ( n - 1 ) 2

1357 - 1 = 2 n - 2

1356 + 2 = 2 n

n = 1358/ 2

n = 679

There is total 679 numbers between 1 to 1357 with difference off 2.