in an ap of 50 term the sum of first + 10 term is 210 s the sum of last 15 term is 2565, find the ap
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10/2 [2a+49d] = 210
2a+49d = 42.............(1)
50/2[2a+49d] - 35/2[2a+34d] = sum of last 15 terms
25[2a+49d] - 35[a+17d] = sum of last 15 terms.
25×42 - 35[a+17d] = 2565
1050 - 35[a+17d] = 2565
35[a+17d] = 1050-2565
a+17d = -1515/35
a+17d = -33/7
please mark it as the brainliest answer
keerthi234:
tq so much
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According to the question:-
Hence required AP is →
3,7,11,15,....,199
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