##S_40 = 3220##
Use the sum formula :
##S_n = n/2 [ 2a + (n-1)d]##
and the ##n##-th term
##T_n = A +(n-1) d##
to solve for ##a## and ##d##.
This will get you
##S_12 = 186## and ##T_20 = 83##.
The equations are
##”i) ” 12/2 [ 2a +11d ] = 186##
and
##”ii) ” a + 19d = 83##
Solve like a linear system :
##{(12a + 66d = 186), (a + 19d = 83) :}##
##a = 83 – 19d##
sub in ##”i)”##
##12 (83 – 19d ) + 66d = 186##
##996 – 228d + 66d = 186##
##162 d = – 810##
##d = -5##
This means that
##a + 19 ( -5 ) = 83##
##a -95 = 83##
##a = 178##
Now
##S_40 = 20 [ 2(178) + 39 ( -5) ]##
## S_40 = 20 [ 356 – 195 ]##
##S_40 = 3220##
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