If the $n t h$ term of AP is $2 n + 5$ . Then find the $A M$ of first and last terms.

99

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98

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100

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44

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Solution

The correct option is D $44$
Given:
$t_{n} = 2 n + 5$
First term $= 2 \times 1 + 5 = 7$
Last term $= 2 \times 38 + 5 = 81$
$∴$ Arithmetic mean $(A . M) . = \frac{7 + 81}{2} = 44$

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