Question

How many terms must of this AP 3,5,7.... must be added to get the simplest 120

Answer 1

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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Answer 2

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