Question

for a sequence of natural number find n if Sn=1540

Answer 1

n 2 + n - 240 = 0. Now this is a quadratic equation, so we will find the factors of this. The factors are (n + 16)(n -15) = 0 ∴ n = -16 or n = 15. But the number of terms never be negative so the value of n = 15. So the sum for first 15 terms of an A.P ...

Answer 2

1,2,3,4,.......... is the sequence
a = 1
d = t2 - t1 = 2-1 = 1
d= 1

Sn = 1540

$sn = \frac{n}{2} (2a + (n -1 )d)$

$1540 = \frac{n}{2}(2(1) + (n - 1)1)$

$1540 = \frac{n}{2} (2 + n - 1)$

$1540 = \frac{n}{2} (n + 1)$

$1540 \times 2 = n(n + 1)$

$3080 = {n}^{2} + n$

${n}^{2} + n - 3080 = 0$

${n}^{2} + 56n - 55n - 3080 = 0$

$n(n + 56) - 55(n + 56) = 0$

$(n + 56)(n - 55) = 0$

$n + 56 = 0 \: \: or \: \: n - 55 = 0$

$n = - 56 \: \: or \: \: n = 55$

But,
n = -56 is not acceptable .

So, n = 55

Thus
$n = 55$
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$hope \: this \: helps$