there are 21 terms in arithmetic progression the sum of last three terms is 123 and the sum of the middle term is 86 find the progression
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n=21 given
T19+T20+T21=123. (last 3 terms in 21 terms)
a+18d+a+19d+a+20d=123
3a+57d=123
÷3
a+19d=41
a=41-19d...(1)
T11=86 (11 is the middle term of 21 terms)
a+10d=86..(2)
put eqn (1) in eqn (2)
41-19d+10d=86
-9d=86-41
-9d=45
d=45/-9
d= -5
put d= -5 in equation (2)
a+10(-5)=86
a=86+50
a=136
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