please solve this question
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t7=49 & t17=289 (given)
t7=a+( n_1 )d
49=a+6d
a+6d=49 ........... (1)
t17 =a+16d
a+16d=289......... (2)
subtract equation 1and 2
a+16d=289
a+6d=49
=10d= 240
d=24
put the place of d=24 in equation 1
a+6×24=49
a=_144+49
a=_95
a=_95 & d=24
t27=n÷2 (2a+(n_1)d)..........formula
t27=27÷2 (2×_95+(27_1)24)
t27= 27÷2 (474)
t27=27×237
t27=6399
therefore sum of t27is 6399
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