# In arithmetic sequence the first term is 2, the common difference is 3. Find the least value of n for which the sum of the first n terms exceeds 1001.

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### 3 Answers

Kindly ignore the last line in the quadratic equation i.e.

`3n^2+155n-154n-2002 = 0`

That was a typing error.

The sum of an arithmetic progression is obtained by S = n/2{2a+(n-1)d}, where n = total number of terms in the progression, a = first term, d = common difference.

Here, a =2, d=3. Least value of n, for which the sum for the first n terms exceeds 1001, has to be determined.

i.e. n/2{2*2 + (n-1)3} > 1001

To solve this inequality, let us first assume that, for n terms the required sum is equal to 1001.

So, n/2{2*2 + (n-1)3} = 1001

Or, n/2(4+3n-3) = 1001

Or, n/2(3n+1) = 1001

or, ` ` `3n^2+n = 2*1001`

or, `3n^2+n-2002 = 0`

or, `3n^2+155n-154n-2002 = 0`

This is a quadratic equation of the type `ax^2+bx+c=0` , the solution of which yields two values of x given by `x = (-b+-sqrt(b^2-4ac))/(2a)`

Upon solving for n, we get the only positive value of n as:

`n=(-1+sqrt(1^2-4*3*(-2002)))/(2*3)`

`=(-1+sqrt(1+24024))/(6)`

`=(-1+155)/(6)`

`=154/6`

=25.67

Returning to the original inequality, n should be an integer, having a value greater than 25.67.

**Therefore, the required value of n is 26.**

`a_1=2 ,d=3 S_n=1001 ,n=?`

`S_n=(n/2){2a+(n-1)d}`

`1001=(n/2)(4+3n-3)`

`2002=4n+3n^2`

`n={-4+-sqrt(16+3xx2002xx4)}/(6)`

`n=(-4+-155)/6`

`n=151/6`

n=26