If the 5th term of an A.P. is 31 and 25th term is 140 more than the 5th term, find the A.P.

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Solution

5th term = 31
a+(5-1)d=31
a+4d=31. ........Eq.1

25th term = 140 + 5th term
a+(25-1)d=140+31
a+24d=171. .......Eq..2

Subtracting Eq2- from Eq 1 we have
a+4d - (a+24d) =31-171
a+4d-a-24d=-140
-20d=-140
or d=7

Put d=7 in Eq 1
a+4d=31
a+4×7=31
a+28=31
a=31-28
a=3

The given A. P. then becomes
a,a+d,a+2d,a+3d,.......
=3,3+7,3+2×7,3+3×7,.........
=3,10,17,24,......

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