question_answer

An arithmetic progression 5, 12, 19...... has 50 terms. Find its last term. Hence find the sum of its last 15 terms.

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 \[l=348\]]

\[=\frac{15}{2}\times 598=4485\]