Question

how many terms of the AP 2,4,6,8 must be added to get the sum 2550​

Answer 1

Hey mate ,

AP = 2,4,6,8

a = 2

d = 4-2 = 2

Sn = 2550

$sn \: = \frac{n}{2} (2a + (n - 1))$

$2550 = \frac{n}{2} (2 \times 2 + (n - 1))$

$5100 = n(4 + (n - 1)$

$5100 = {n}^{2} + 3n$

$0 = {n}^{2} + 3n - 5100$

$d = {b}^{2} - 4ac$

$d = 20409$

TERMS CAN'T BE DEFINED

Answer 2

Answer:

see here

Step-by-step explanation:

a=2

d= 2

Sn= 2550

n=?

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

2550=n/2(4+(n-1)2)

2550x2= n(4 - 2 +2n)

5100=2n+2n²

2n²+2n-5100=0

2(n²+n-2550)=0

n²+n-2550=0

n²-50n+51n-2550=0

n(n-50)+51(n-50)=0

(n+51)(n-50)=0

hence n is equal to 50