Question

how many terms of the AP 7, 14, 21... are needed to give the sum of 5740​

Answer 1

Answer:

Step-by-step explanation:

a= 7, d= 14-7 = 7, sn=5740

Sn= n/2(2a+(n-1)d

5740 x 2 = n( 2 x7 +(n-1)7)

11480 = n [ 14+7n -7) ]

11480= n (7n -7)

11480 =7n^2 -7n

7n^2 -7n - 11480 = 0

Dividing by 7

n^2 +n + 1640=0

By using quadratic formula

We get, n= 40

Answer 2

Answer:

40

Step-by-step explanation:

Here, a = 7

d = 14 - 7 = 7

Let the term of the AP needed to give the sum be n

We know that, sum of A.P (S) = $\frac{n}{2}(2a + (n - 1)d)$

Here, S = 5740

Substituting the values, we get

$5740 = \frac{n}{2}(2\times 7 + (n - 1)7)\\ \\5740 \times 2 = n(14 + 7n - 7)\\\\11480 = n(7 + 7n)\\\\11480 = 7n^{2} + 7n$

→ 7n² + 7n - 11480 = 0

→7(n² + n - 1640) = 0

→ n² + n - 1640 = 0

→ n² + 41n - 40n - 1640 = 0

→ n(n + 41) - 40(n + 41) = 0

→(n + 41)(n - 40) = 0

Hence, n = - 41 or n = 40

Since, term can not be negative, n = 40