Question

the sum of the first three terms of an AP is 33 if the first term and third term exceeds the 2nd term by 29. find AP and also find the sum of first 25 terms
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Answer 1

Step-by-step explanation:

$let \: the \: series \: of \: ap = (a - d). : \: a \: :( a + d) \\ \ sum \: of \: the \: three \: terms \: are \: a - d + a + a + d \: = 3a \: \\ 3a = 33 \\ a = 11 \: \\ from \: the \: given \: condition \: is \: \\ (a + d) \: (a - d) = a + 29 \\ \ (a + d) \: ( a - d) = 11 + 29 = 40 \\ \\ { {a}^{2} - {d}^{2} }^{} = 40 \\ 11 {}^{2} - 40 = {d}^{2} \\ \\ 121 - 40 = {d}^{2} \\ \\ {d }^{2} = 81 \\ d = \sqrt{81 } \\ \\ d = + - 9 \\ \\ thus \: the \: ap \: is \: 2 \: \: \: \: 11 \: \: \: 19 \\ \\ please \: mark \: me \: as \: brainliest$

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