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Question

Find the 31st term of an A. P. whose 10th term is 38 and 16th term is 74.

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Solution

Let first term of AP is a and common difference is d.
Use formula of tn=a+(n1)d.......(i)
Given, 10th term is 38 and 16th term is 74.
So, substitute n=10 and 16 in equation(i), we get
t10=a+(101)d=38 and t16=a+(161)d=74
a+9d=38 and a+15d=74
Substitute a=389d in a+15d=74, we get
389d+15d=74
38+6d=74
6d=7438
d=366
d=6
Substitute d=6 in a=389d, we get
a=389(6)
a=3854
a=16
Thus, 31th term is t31=a+30d
Substitute, a=16 and d=6 in above expression.
t31=16+30(6)
=16+180
t31=164


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