Question

If the first term and the last term of a finite A.P. are 5 and 95 respectively and \[d = 5\] , find n and \[{S_n}\] .

Answer 1

Hint: For finding n, use the formula \[{T_n} = a + \left( {n - 1} \right)d\] . After finding n, find \[{S_n}\] using the formula \[{S_n} = \dfrac{n}{2}\left( {a + l} \right)\] .

Complete step-by-step answer:
It is given that the first term of an A.P. \[a = 5\] , last term \[l = 95\] and difference between terms \[d = 5\] .
Now, first we have to find \[n\] and for that we have to use the formula \[{T_n} = a + \left( {n - 1} \right)d\] .
 \[
  \therefore {T_n} = a + \left( {n - 1} \right)d \\
  \therefore 95 = 5 + \left( {n - 1} \right) \times 5 \\
  \therefore 95 = 5 + 5n - 5 \\
  \therefore 5n = 95 \\
  \therefore n = \dfrac{{95}}{5} \\
  \therefore n = 19 \\
 \]
Thus, we get a number of terms in the given A.P. as \[n = 19\] .
Now, for the sum of the terms, we will use the formula \[{S_n} = \dfrac{n}{2}\left( {a + l} \right)\] .
 \[{S_n} = \dfrac{n}{2}\left( {a + l} \right)\]
      \[
   = \dfrac{{19}}{2}\left( {5 + 95} \right) \\
   = \dfrac{{19}}{2}\left( {100} \right) \\
   = 19 \times 50 \\
   = 950 \\
 \]
Thus, we get the sum of the terms of the given A.P. as \[{S_n} = 950\] .

Note: Alternate method for sum:
Here, we can also use the formula \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\] for the sum of terms of given A.P.
 \[\therefore {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
          \[
   = \dfrac{{19}}{2}\left[ {2 \times 5 + \left( {19 - 1} \right) \times 5} \right] \\
   = \dfrac{{19}}{2}\left( {10 + 18 \times 5} \right) \\
   = \dfrac{{19}}{2}\left( {10 + 90} \right) \\
   = \dfrac{{19}}{2}\left( {100} \right) \\
   = 19 \times 50 \\
   = 950 \\
 \]