Accountancy, asked by snehakataria419, 17 days ago

An arithmetic progression has t1 = 7 , tn = 77 , and , Sn = 420 . Find n and d.​

Answers

Answered by saifulaqib7602
1

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

-----------

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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