Math, asked by 9964048059, 11 months ago

In an ap 1+4+7+10+ .........+X=287,then find the value of X

Answers

Answered by sanketj
15

In the A.P.

1, 4, 7, 10, ..., x

first term, (a) = 1

common difference, (d) = 3

 a_{n} = x \\ \\ S_{n} = 287

287 =  \frac{n}{2} (2a + (n - 1)d) \\ 287 =  \frac{n}{2} (2(1) + (n - 1)(3)) \\ 287  \times 2 = n(2 + 3n - 3) \\ 574 = n(3n - 1) \\ 574 = 3 {n}^{2}  - n \\ 3 {n}^{2}  - n - 574 = 0 \\ 3 {n}^{2}  - 42n  + 41n - 574 = 0 \\ 3n(n - 14) + 41(n - 14) = 0 \\ (3n + 41)(n - 14) = 0 \\  \\ 3n + 41 = 0 \:  \:  \:  \:  \:  \:  \:  \: or \:  \:  \:  \:  \:  \:  \:  \: n - 14 = 0 \\  \:  \:   \:  \:  \:  \:  \: \:  \:  \:  \: n =  -  \frac{41}{3}  \:  \: or \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  n = 14

but 'n' can't be negative or a fraction

hence, n = 14

 a_{n}  = x \\  a_{14}  = x \\ x = a + 13d \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   ... \:  a_{n}  = a + (n - 1)d \\ x = 1 + 13(3) \\ x = 40

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