Question

find the number of terms the AP 2 + 4 + 6.......= 420​

Answer 1

Sn=420

a=2

d=2

Sn=n/2(2a+(n-1)d)

420=n/2(2(2)+(n-1)d)

420=n/2(4+(nd-d))

420 X 2=n(4+nd-d)

840=4n-n²d-d

PUT d=2

840=4n-2n²-2

840=-(2n²-4n+2)

840=-(2n²-2n-2n+2)

840=-(2n(n-1)-2(n-1))

840=((2n-2)(n-1))

THEREFORE

n=421 or n=841

Answer 2

Answer:

20

Step-by-step explanation:

Let n be the number of terms

Then...

2 + 4 + ... + 2n = 420

=> 2 × ( 1 + 2 + ... + n ) = 420

=> 2 × n ( n + 1 ) / 2 = 420

=> n ( n + 1 ) = 420

=> n² + n = 420

=> 4n² + 4n = 4×420 = 1680

=> (2n)² + 2(2n) + 1 = 1681

=> (2n + 1)² = 1681 = 41²

=> 2n+1 = ±41

=> 2n = -1 ± 41

=> 2n = -1 + 41 = 40   ( since n must be even )

=> n = 20