How many terms must of this AP 3,5,7.... must be added to get the simplest 120
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Answered by
2
Answer:
Step-by-step explanation:
Sn = n/2[2a+(n-1)d]
given,Sn=120,a=3 , d=2
120=n/2[2×3+(n-1)2]
120×2=n[6+2n-2]
240=n[4+2n]
240=4n+2n²
240=2(2n+n²)
120=n²+2n
0=n²+2n-120
0=n²+12n-10n-120
0=n(n+12)-10(n+12)
0=(n+12)(n-10)
n=-12 , 10
hence no of terms can't be negative
so n=10
thus there are 10 terms of the given AP must be added to get sum 120.
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Answered by
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Answer:
10
Step-by-step explanation:
a=3 d=5-3=2 Sn=120
Sn=(n/2){2a+[n-1]d}
120=(n/2){2*3+[n-1]2}
120=(n/2){6+2n-2}
120=(n/2){4+2n}
120=n{2+n}
120=2n+n^2
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)=0 (n+12)=0
n=10 n=-12(rejected because terms cannot be negative)
Therefore 10 terms must be added to get 120
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