How many terms of the AP 2,4,6,8,10..... are needed the sum 210?
Answers
Answered by
5
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.
Attachments:
Answered by
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!!!!....
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