Question

How many terms of the AP 2,4,6,8 are needed to get sum 210

Answer 1

HEY MATE!!
The answer to the question is:
a=2
d=2
Sn=210
n=?
Sn=n/2 × [2a+(n-1)d]
210=n/2×[4+(n-1)2]
210×2=n[4+(n-1)2]
420=4n+(n-1)2n
420=4n+2n^2-2n
420=2(n^2-n)
210=n^2-n
n^2-n-210=0
n^2-15n+14n-210=0
n(n-15)+14(n-15)=0
(n+14)(n-15)=0
n+14=0. n-15=0
n=-14. n=15
therefore n=15
=>15 terms are needed to get their sum as 210
Please mark BRAINLIEST if satisfied....
HOPE IT HELPS YOU!!!!....

Answer 2

AP is 2,4,6,8.
first term a =2
common difference d =4-2 =2

sum of n terms of AP is Sn,
so
Sn = n/2 *( 2a + (n-1) * d)

because sum is given 210 and we need to find n.
So 210=n/2 *(2*2 +(n-1)*2)

=> 210*2 = n *(4 +2n -2)
=> 420 =n *(2+2n)
=> 420/2 = n + n*n
=> 210 = n*n +n
=> n*n + n -210 = 0
Quadratic equation
=> n*n +15n -14n -210 =0

=> n(n +15) -14(n +15) =0
=> (n +15)(n -14)=0
=> n = 14 or n = -15
n can not be negative because it is number of terms.
Hence. n = 14.

So 14 terms needed to get sum of AP 210.