Question

Q.28 Find the maximum sum of the A.P: 120, 116, 112, ……….​

Answer 1

Answer:

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here is your answer dude mark as brainlist

Answer 2

$S_{max} = 2460$

Step-by-step explanation:

120, 116, 112, ........ is a decreasing A.P. series and its first term is 120 and the common difference is - 4.

So, the sum of this A.P. series will be maximum when we add the positive terms of the A.P.

Therefore, $S_{max} = 120 + 116 + 112 + 108 + .............. + 0$

⇒ $S_{max} = \frac{41}{2}(120 + 0) = 41 \times 60 = 2460$ (Answer)

{Since the number of terms is 41, and the first term and last terms are 120 and 0 respectively}