Question

how many terms of an AP 3,5,7,... must be taken so that the sum is 120?

Answer 1

hope u understand mark is as brainliest

Answer 2

Answer:

10 terms

Step-by-step explanation:

A.P.= 3,5,7.......

a = first term = 3

d = common difference = 2

Sum of first n terms in A.P. = $S_n=\frac{n}{2} (2a+(n-1)d)$

Now we are given that the sum of n terms is 120

So, to find n :

$120=\frac{n}{2} (2(3)+(n-1)2)$

$240=n(6+2n-2)$

$240=4n+2n^2$

$2n^2+4n-240=0$

$n^2+2n-120=0$

$n^2+12n-10n-120=0$

$n(n+12)-10(n+12)=0$

$(n+12)(n-10)=0$

$n=-12,10$

Since number of terms cannot be negative So, n = 10

Thus 10 terms of AP must be taken so that the sum is 120