The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41st term of that A.P.
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Answer:
a11-16
a+10b=16....equation1
a21-29
a+20b=29....equation2
subtracting 1 and 2
a+1b=16
-a-+21b=-21
=-11b=-15
b=15\11
put
b=15\11 in 1
a+110\15=16
15a+110=240
15a=240-110
a=130\15=26\3
a31=26\3+30(15/11)
=26/3+450/11
286+135/33
=421/33
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