How many terms of the AP 3, 5, 7, 9, ... must be added to get the
sum 120?
Answers
Answered by
6
Answer:
hope it's helpful to you
Step-by-step explanation:
120=
2
n
[a+a+(n−1)d]
where n= no of terms
a=first term
d=common difference
120=
2
n
[3+3+(n−1)2]
240=n[3+3+(n−1)2]
240=n[6+2n−2]
240=n[4+2n]
n
2
+2n−120=0
n
2
+12n−10n−120=0
n(n+12)−10(n+12)=0
(n−10)(n+12)=0
n=10
no of terms must be added 10−4=6 terms
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