The 3rd term of an arithmetic sequence is 43 and 6th term 76 write first 3 term and find the 10 th term
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Answer:
A3=43
A6=76
a+2d=43
a+5d=76
a6-a3
3d=33
d=11
a+2×11=43
a=43-22
a=21
substitute a and d in a , a+d and a+2d
a+9d=21+9×11
21+99=120
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