100 A.Ms are inserted between 20 and 80. Find the sum of first A.M and last A.M
A
100
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B
80
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C
60
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D
92
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Solution
The correct option is A
100
Let the common difference be d. Then the A.P will be 20,20+d–––––––,20+2d......80−2d,80−d–––––––,80 ⇒ Sum of first and last A.M (underlined terms) ⇒20+d+80−d=100