If the 5th term 12 term of its Ap are 30 and 65 and find the sum of its 31st term
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ATQ,
a5= a+4d=30 --------1
a12=a+11d=65 --------2
By elimination method ,
Subtracting eq 1 from eq 2
a+11d=65
-(a+4d=30)
________
7d =35
________
d=5
Putting the value of d in eq 1.
a+4d=30
a=30-20
a=10
S 31= n/2[2a+(n-1)d]
= 31/2[20+34×5]
= 31/2[190]
= 31×95
= 2945
Sum of 31 st term of AP is 2945.
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