Question

how many terms of an a.p. 1,4,7,... are needed to give sum 2380

Answer 1

Answer:

40 terms of an a.p. 1,4,7,--------- are needed to give sum 2380.

Step-by-step explanation:

Given AP 1 , 4 , 7 , -----

First-term (a) = 1 and

And the common difference (d) = second term - first term

= 4-1  = 3

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

As Sn = 2380

So

$(S_n) = \frac{n}{2} [2a + (n-1)d]\\\frac{n}{2} [2(1) + (n-1)3] = 2380\\\frac{n}{2} [2+ 3n- 3] = 2380\\\frac{n}{2} [ 3n- 1] = 2380\\n ( 3n - 1) = 2 \times 2380\\3n^2 - n = 4760\\3n^2 - n - 4760= 0\\3n^2 - 120n + 199n - 4760= 0\\n = 40 and n = -119/3$

The number of terms can never be negative, so n = 40

Answer 2

40 terms

The given AP is 1,4,7, . . . .  . .

The first term of this AP is a=1

And the common difference d=3

Sum of n terms in AP, S = n/2[2a + (n − 1) × d]

Here, n= nth term

The sum of n term is given that is equal to 2380

2380= n/2[2×1 + (n − 1) × 3]

After solving the equation we will get the value of n=40(ANSWER)

So, the total number of terms required is equal to 40 terms.