Question

How many terms of the AP 2,4,6,8,10..... are needed the sum 210?​

Answer 1

In the attachment I have answered this problem.

Concept:

Sum of n terms of an AP

= n/2[2a+(n-1)d]

From the attachment it is clear that 14 terms are needed to get the sum 210.

See the attachment for detailed solution.

Answer 2

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!!!!....