(iv) given a = 15, S=125, find d and a
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Answer:
a=15
Sum of ten terms=125
n=10
Sn=n/2[2a+(n-1)d]
125=10/2[2×15+(10-1)d]
125=5[30+9d]
125/5=30+9d
25=30+9d
25-30=9d
-5=9d
d=-5/9
a10=a+(n-1)d
=15+(10-1)×-5/9
=15+9×-5/9
=15-5=10
last term=10
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