If the sum of the consecutive integers from –42
[#permalink]
07 Aug 2021, 19:59
Consecutive integers from \(-42\) to \(n\), the sum is \(372\)
When there are consecutive integers then the sum of the numbers between \(-42\) to \(42\) will be zero.
From \(43, 44, 45,...\) the sum of the series will be \(372\).
The sum of the series, when last digit is known = \(S_n =\) \(\frac{n}{2} (2a + (n-1)d)\)
\(a =\) first term & \(d =\) common difference
\(372 = \frac{n}{2} (2(43) + (n-1))\)
By solving this, the answer of \(n\) will be \(8\), which means that the no will be \(50\)
OR
you can try adding nos \(43, 44, 45,...\) up until the sum is \(372\) and the last no will be \(50\)
Answer D