Question

In an A.P, the first term is 1 and the common difference is 4. How many terms of the A.P must be taken for their sum to be equal to 120?

Answer 1

8 terms of the A.P must be taken for their sum to be equal to 120.

Step-by-step explanation:

Given : In an A.P, the first term is 1 and the common difference is 4.

To find : How many terms of the A.P must be taken for their sum to be equal to 120?

Solution :

The sum of n numbers of A.P is

$S_n=\frac{n}{2}[2a+(n-1)d]$

Where, a is the first term a=1

d is the common difference d=4

$S_n=120$ is the sum

Substitute the values,

$120=\frac{n}{2}[2(1)+(n-1)4]$

$240=n[2+4n-4]$

$240=4n^2-2n$

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

$(n-8)(2n+15)=0$

$n=8,-\frac{15}{2}$

Reject $n=-\frac{15}{2}$

The value of n is 8.

So, 8 terms of the A.P must be taken for their sum to be equal to 120.

#Learn more

(a) (b)

5 3 (d) 6

7. How many terms of an AP must be taken for their sum to be equal to 120 if its third term is 9 and

the difference between the seventh and second term is 20 ?

(a) 7 (b) 8 (c) 9 (d) 6

https://brainly.in/question/11808334

Answer 2

The total number of terms is 8.

Step-by-step explanation:

Here, first term (a) = 1, common difference (d) = 4 and

$S_{n} =120$

Let n be the number of terms.

To find, the total number of terms = ?

We know that,

The sum of nth terms,

$S_{n} =\dfrac{n}{2} [2a+(n-1)d]$

∴ $\dfrac{n}{2} [2(1)+(n-1)4]$ = 120

⇒ n(2n -1) = 120

⇒ 2$n^2$- n - 120 = 0

⇒ 2$n^2$- 16n + 15n - 120 = 0

⇒ 2n(n - 8) + 15(n - 8) = 0

⇒  (n - 8)(2n + 15) = 0

⇒ n = 8 [∵ n is never be fraction]

Thus, the total number of terms is 8.