Multiple choice

The first term of an $A.P.$ is $10$ and its $41^{st}$ term is $81$. Find the sum of the series, if $81$ is the last term.

  1. $1025$
  2. $1865.5$
  3. $858$
  4. $962$
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B Correct answer
Explanation

Sum = (n/2) * (first + last). n = 41. Sum = (41/2) * (10 + 81) = 41 * 45.5 = 1865.5.

AI explanation

Using the arithmetic progression formula for the nth term, T_n = a + (n - 1)d, we substitute the given values to find the common difference. We have 81 = 10 + 40d, which yields 40d = 71 and d = 1.775. Because 81 is the last term, there are 41 terms in the series. The sum of the series is calculated using S_n = n/2 times (first term + last term), so S_41 = 41/2 times (10 + 81). This equals 20.5 times 91, which results in 1865.5.

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