given a = 15, S10 = 125, find d and a10
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a3= 15. a+2d=15.also s10= 125. 10/2(2a+(10-1)d)=125. 5(2a+9d)=125. 2a+9d=25.by subtracting eq (I ) and ( ii) from it we get. 2(a+2d)-(2a+9d)=2×15×-25. 4d-9d=30-25. -5d =5. d=-5/5=-1 . now a10=a+9d. =[a+2d]+7d. =15+7(-1). 15-7=8 . d=-1 and a10 =8. ok
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