Question

Sum of how many terms of the AP 1, 4, 7, 10,...... will be 715?

Answer 1

Step-by-step explanation:

Given series is:

1 + 4+ 7 + 10 +... = 715

Here, a = 1, d = 3, Sn = 715,

We need to find: n

$\because \: S_n = \frac{n}{2} \{2a + (n - 1)d \} \\ \\ \therefore \: 715 = \frac{n}{2} \{2 \times 1 + (n - 1) \times 3 \} \\ \\ \therefore \: 715 \times 2 = n \{2 +3n - 3 \} \\ \\ \therefore \: 1430 = n \{3n - 1 \} \\ \\ \therefore \: 3 {n}^{2} - n - 1430 = 0\\ \\ \therefore \: 3 {n}^{2} - 66n + 65n- 1430 = 0\\ \\ \therefore \: 3n(n - 22) + 65(n - 22) = 0 \\ \\ \therefore \: (n - 22)(3n + 65) = 0 \\ \\ \therefore \: (n - 22) = 0 \: \: or \: \: (3n + 65) = 0 \\ \\ \therefore \: n = 22 \: \: or \: \: n = - \frac{65}{3} \\ \\ \: \: \: \: \: but \: n \neq \: - \frac{65}{3} \\ \\ \huge \red {\boxed{ \boxed{\therefore \: n = 22}}}$

Hence, sum of 22 terms will be 715.