ARITHMETIIC PROGRESSIONS
Asked by MANOJKUMAR | 15th May, 2010, 01:33: PM
S11 = 11(2a0 + 10d)/2 = 17
2a0 + 10d = 34/11 ... (1)
S17 = 17(2a0 + 16d)/2 = 11
2a0 + 16d = 22/17 ... (2)
Solving (1) and (2) for a0 and d,
d = -56/187 and a0 = 2899/2057.
S28 = 28(2(2899/2057) + 27(-56/187))/2
Regards,
Team,
TopperLearning.
Answered by | 17th May, 2010, 11:40: AM
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