Question

How many terms of the A. P. 3,5,7,9.....must be added to get the sum 120?

Answer 1

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:

number of terms will be 10

apply formula :

$sum = \frac{n}{2} (2a + (n - 1)d$

where d = 2(5-3 =2)

sum = 120(given)

a =3