Question

how many terms of A.P 3,5,7,9... must be taken to get the sum 120​

Answer 1

Answer:

Ap = 3+5+7+9=24

Step-by-step explanation:

=120÷24=5

Answer 2

Step-by-step explanation:

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

Sn = n/2[2a+(n-1)d]given, Sn=120,a=3 , d=2 120=n/2[2x3+(n-1)2] 120x2=n[6+2n-2]

Sn = n/2[2a+(n-1)d]given, Sn=120,a=3 , d=2 120=n/2[2x3+(n-1)2] 120x2=n[6+2n-2]240=n[4+2n]

Sn = n/2[2a+(n-1)d]given, Sn=120,a=3 , d=2 120=n/2[2x3+(n-1)2] 120x2=n[6+2n-2]240=n[4+2n]240=4n+2n?

Sn = n/2[2a+(n-1)d]given, Sn=120,a=3 , d=2 120=n/2[2x3+(n-1)2] 120x2=n[6+2n-2]240=n[4+2n]240=4n+2n?240=2(2n+n?)

Sn = n/2[2a+(n-1)d]given, Sn=120,a=3 , d=2 120=n/2[2x3+(n-1)2] 120x2=n[6+2n-2]240=n[4+2n]240=4n+2n?240=2(2n+n?)120=n²+2n

Sn = n/2[2a+(n-1)d]given, Sn=120,a=3 , d=2 120=n/2[2x3+(n-1)2] 120x2=n[6+2n-2]240=n[4+2n]240=4n+2n?240=2(2n+n?)120=n²+2nO=n²+2n-120

Sn = n/2[2a+(n-1)d]given, Sn=120,a=3 , d=2 120=n/2[2x3+(n-1)2] 120x2=n[6+2n-2]240=n[4+2n]240=4n+2n?240=2(2n+n?)120=n²+2nO=n²+2n-120O=n2+12n-10n-120

Sn = n/2[2a+(n-1)d]given, Sn=120,a=3 , d=2 120=n/2[2x3+(n-1)2] 120x2=n[6+2n-2]240=n[4+2n]240=4n+2n?240=2(2n+n?)120=n²+2nO=n²+2n-120O=n2+12n-10n-120O=n(n+12)-10(n+12)

Sn = n/2[2a+(n-1)d]given, Sn=120,a=3 , d=2 120=n/2[2x3+(n-1)2] 120x2=n[6+2n-2]240=n[4+2n]240=4n+2n?240=2(2n+n?)120=n²+2nO=n²+2n-120O=n2+12n-10n-120O=n(n+12)-10(n+12)O=(n+12)(n-10)

Sn = n/2[2a+(n-1)d]given, Sn=120,a=3 , d=2 120=n/2[2x3+(n-1)2] 120x2=n[6+2n-2]240=n[4+2n]240=4n+2n?240=2(2n+n?)120=n²+2nO=n²+2n-120O=n2+12n-10n-120O=n(n+12)-10(n+12)O=(n+12)(n-10)n=-12, 10

Sn = n/2[2a+(n-1)d]given, Sn=120,a=3 , d=2 120=n/2[2x3+(n-1)2] 120x2=n[6+2n-2]240=n[4+2n]240=4n+2n?240=2(2n+n?)120=n²+2nO=n²+2n-120O=n2+12n-10n-120O=n(n+12)-10(n+12)O=(n+12)(n-10)n=-12, 10hence no of terms can't be negative

Sn = n/2[2a+(n-1)d]given, Sn=120,a=3 , d=2 120=n/2[2x3+(n-1)2] 120x2=n[6+2n-2]240=n[4+2n]240=4n+2n?240=2(2n+n?)120=n²+2nO=n²+2n-120O=n2+12n-10n-120O=n(n+12)-10(n+12)O=(n+12)(n-10)n=-12, 10hence no of terms can't be negativeso n=10

Sn = n/2[2a+(n-1)d]given, Sn=120,a=3 , d=2 120=n/2[2x3+(n-1)2] 120x2=n[6+2n-2]240=n[4+2n]240=4n+2n?240=2(2n+n?)120=n²+2nO=n²+2n-120O=n2+12n-10n-120O=n(n+12)-10(n+12)O=(n+12)(n-10)n=-12, 10hence no of terms can't be negativeso n=10thus there are 10 terms of the given AP must be added to get sum 120.

Sn = n/2[2a+(n-1)d]given, Sn=120,a=3 , d=2 120=n/2[2x3+(n-1)2] 120x2=n[6+2n-2]240=n[4+2n]240=4n+2n?240=2(2n+n?)120=n²+2nO=n²+2n-120O=n2+12n-10n-120O=n(n+12)-10(n+12)O=(n+12)(n-10)n=-12, 10hence no of terms can't be negativeso n=10thus there are 10 terms of the given AP must be added to get sum 120.hope this helps!

Sn = n/2[2a+(n-1)d]given, Sn=120,a=3 , d=2 120=n/2[2x3+(n-1)2] 120x2=n[6+2n-2]240=n[4+2n]240=4n+2n?240=2(2n+n?)120=n²+2nO=n²+2n-120O=n2+12n-10n-120O=n(n+12)-10(n+12)O=(n+12)(n-10)n=-12, 10hence no of terms can't be negativeso n=10thus there are 10 terms of the given AP must be added to get sum 120.hope this helps!Please mark it as brainliest....