An arithmetic progression has t1 = 7 , tn = 77 , and , Sn = 420 . Find n and d.
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Answer:
n=70
t=11/10 or 0.1
s=6
Explanation:
eq1 t1=7 eq2 TN=77
subtract TN from t1
TN =77
-
T1=7
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n=70
put n=70 in eq1
t(70)=77
divide both side by 70
t=10/11
now put the value of n in Sn=420
s(70)=420
divide both side by 70
s70/70=420/70
s=6
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