Question

How many terms of the A.P 2,4,6.......must be taken so that their sum is 156?

Answer 1

Let n terms should be taken

Then, Sn = 156
           n/2 [2*2 + (n-1)2] = 156
       so, n[4-2+2n] = 312  
             n[2 + 2n] = 312
       so, 2n² + 2n - 312 = 0
         so, n² + n - 156 = 0
          n² +13n - 12n - 156 = 0
       n(n+13) -12(n+13) = 0
      so, n = 12, -13

But n can'be negative

Hence n = 12
So, Required terms whose sum is 156 is 12 terms

Answer 2

Given,
a = 2
d = 2
Sn= 156
n=?

Sn = n/2[2a+(n-1)×d]
=> 156= n/2[2×2 +(n-1)×2]
=> 156×2 = n[4+2n-2]
=> 312 = 4n+2n^2 - 2n
=> 312= 2n+2n^2
=> 2n^2 + 2n +312=0
=> n= - 13 Ans