Question

If the sum of $7$ terms of an A.P is $49$ and that of $17$ terms is $289$, find the sum of n terms.
A.$n$
B.${n^2}$
C.${n^3}$
D.${n^4}$

Answer 1

Hint: Use the formula of A.P. for $n = 7$ and $n = 17$ to get the value of the first term and the common difference and calculate the sum of n terms

Complete step-by-step answer:

We know that the formula for the sum of n terms of Arithmetic Progression as
\[S_n^{} = {\text{ }}\dfrac{n}{2}[2a + (n - 1)d]\]
 In this formula $a = $first term of A.P. , $n = $number of terms in A.P. , $d = $common difference of A.P.,
 $S_n^{} = $sum of $n$terms
 Now we should substitute the value of n as $7$ and sum as $49$ in the above formula
 and get

 \[S_7^{} = \dfrac{7}{2}[2a + (7 - 1)d] = 49\]
  Now on solving this equation we get as
  $\dfrac{{14 - 6d}}{2} = a$
  Now forming the equation we get
  \[ \Rightarrow a = 7 - 3d..........\left( 1 \right)\]
   Now on substituting the value of n as $17$and sum as $289$in the formula of Arithmetic Progression we
   get
   \[S_{17}^{} = \dfrac{{17}}{2}[2a + (17 - 1)d] = 289\]
    Now we solve this equation and get

 $ \dfrac{{34 - 16d}}{2} = a$
  $ \Rightarrow a = 17 - 8d........(2) \\ $

     Now equating equation 1 and 2 we get
     d=2 and then putting in equation 1 we get
     a=1
     now again substituting the value of a and d in formula of sum of A.P. and we get
     \[
  S_n^{} = {\text{ }}\dfrac{n}{2}[2a + (n - 1)d] \\
  {\text{ = }}\dfrac{n}{2}[2(1) + (n - 1)2] \\
  {\text{ = }}\dfrac{n}{2}(2n) \\
   \Rightarrow S_n^{} = {n^2} \\
 \]
     Hence answer is option B

Note: while solving questions of A.P. we should use the formula of the general sum and n term and then calculate the value of first term and common difference and then calculate the sum and get the result.