Question

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

Answer 1

Answer:

10 terms

thus there are 10 terms of the given AP must be added to get sum 120.

Answer 2

Answer

According to the question,

First term = 3

Common difference = (5 - 3) = 2

Let the required number of terms be n,

$\sf \dashrightarrow {S}_{n} = 120$

$\sf \dashrightarrow \dfrac{n}{2} . {2a + (n - 1) d} = 120$

$\sf \dashrightarrow \dfrac{n}{2} . {2 \times 3 + (n - 1) \times 2} = 120$

$\sf \dashrightarrow \dfrac{n}{2} . (2n + 4) = 120$

$\sf \dashrightarrow {n}^{2} + 2n - 120 = 0$

$\sf \dashrightarrow {n}^{2} + 12n - 10n - 120 = 0$

$\sf \dashrightarrow n (n + 12) - 10 (n + 12) = 0$

$\sf \dashrightarrow (n + 12) (n - 10) = 0$

$\sf \dashrightarrow n + 12 = 0 \: or \: n - 10 = 0$

$\sf \dashrightarrow n = -12 \: or \: n = 10$

The number of terms cannot be negative. So,

$\sf \dashrightarrow n = 10$

Thus, the required number of terms is 10.