If 1+5+7+10+......+x = 287 find the …
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Yogita Ingle 6 years, 10 months ago
Given, 1 + 4 + 7 + 10 +...+ x = 287
This series in AP.
Now, first term a = 1, common difference d = 4 - 1 = 3
Last term l = x
Let number of terms = n
Now, Sum of the seires Sn = 287
=> (n/2) ×{2a + (n - 1)d} = 287
=> (n/2) ×{2 + (n - 1)3} = 287
=> (n/2) ×{2 + 3n - 3} = 287
=> (n/2) ×{3n - 1} = 287
=> n ×(3n - 1) = 287 ×2
=> 3n2 - n = 574
=> 3n2 - n - 574 = 0
=> (n - 14) ×(3n + 41) = 0
=> n = 14, -41/3
Since n can not be negative,
So, n = 14
i.e. there are 14 terms in the series and x is the 14th term.
So, x = a14 = a + (14 - 1)d
=> x = a + 13d
=> x = 1 + 13 ×3
=> x = 1 + 39
=> x = 40
So, the value of x is 40
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