determine the AP whose third term is 60 and the 7 term exceeds the 5th term by 12
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n=3 and Tn=60.
now from formula a+(n-1)d=Tn
so a+(3-1)d=60
a+2d=60....Eq.... (i)
again, from question
7 term exceeds 5th term by 12.
therefore, a+6d-(a+4d)=12
a+6d-a-4d=12
2d=12
d=6
now, putting d=6 in Eq.. (i)
a+2×6=60
a+12=60
a=60-12
a=48
since,AP= a,a+d,a+2d,...........
so AP=48, 54, 60, ........answer
now from formula a+(n-1)d=Tn
so a+(3-1)d=60
a+2d=60....Eq.... (i)
again, from question
7 term exceeds 5th term by 12.
therefore, a+6d-(a+4d)=12
a+6d-a-4d=12
2d=12
d=6
now, putting d=6 in Eq.. (i)
a+2×6=60
a+12=60
a=60-12
a=48
since,AP= a,a+d,a+2d,...........
so AP=48, 54, 60, ........answer
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