Question

how many terms of ap3579 must be taken to get the sum120

Answer 1

HOPE IT WILL HELP YOU:

Answer 2

Number of terms will be 10

Let the number of terms in the progression be n.

Given,

• A.P = 3, 5, 7, 9, .........
• Common difference (d) = 2
• Sum (S) = 120
• First term (a) = 3
• Number of terms (n) = n

Here,

We have to apply the formula :  

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

Applying the values

,

$120 = \frac{n}{2} [(2\times3) + (n - 1)2]$

⇒ (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

Here,

Number of terms cannot be negative

∴ n = 10

Thus,

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