Answer:
Given, AP is 5, 12, 19 ............. Here, \[n=50,\,\,a=5,\,\,d=12-5=19-12=7\] Now, \[{{T}_{50}}=a+(50-1)d\] \[\Rightarrow {{T}_{50}}=5+(49)7=348\] 15 terms from last \[=(50-15+1)\] terms from starting \[{{T}_{36}}=a+(36-1)d\] \[=5+35(7)\] \[=250\] \[\therefore \] Sum of last 15 terms \[=\frac{n}{2}(a+l)\] \[=\frac{15}{2}(250+348)\] [\[\because \,\,a=250\] and \[l=348\]] \[=\frac{15}{2}\times 598=4485\]
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