Question

35. If the sum of first 7 terms of AP is 49 and that of it first 17 terms is 289. Find the sum of first n terms of AP.​

Answer 1

Answer:

The sum of first n terms of AP is $n^{2}$.

Step-by-step explanation:

Given that the sum of the first 7 terms of AP is 49 and the first 17 terms are 289.

The formula to find the first n terms of an AP is given as,

      $S_{n}=\dfrac{n}{2}\left( 2a+\left( n-1\right) d\right)$

where 'n' is the number of terms, 'a' is the first term, and 'd' is the common difference. To find the first n terms of AP we want to find the values of 'a' and 'd'.

The sum of the first 7 terms ⇒

      $\begin{array}{l}S_7=\dfrac{7}{2}\left(2a+\left(7-1\right)d\right)=49\\\\\dfrac{7}{2}\left(2a+6d\right)=49\\\\\dfrac{7}{2}\left(a+3d\right)2=49\\\\7\left(a+3d\right)=49\\\\a+3d=\dfrac{49}{7}=7\\\end{array}$→ (1)

Sum of the first 17 terms ⇒

                    $\begin{array}{l}S_n=\dfrac{17}{2}\left(2a+\left(17-1\right)d\right)=289\\\\\dfrac{17}{2}\left(2a+16d\right)=289\\\\\dfrac{17}{2}\left(a+8d\right)2=289\\\\17\left(a+8d\right)=289\\\\a+8d=\dfrac{289}{17}=17\\\end{array}$→ (2)

Now subtract equation (2) from (1) ⇒

                      $a+8d-(a+3d)=17-7\\\\a+8d-a-3d=10\\\\5d=10\\\\d=2$

Now we have found the value of 'd'. To find 'a', substitute d=2 in equation (1) or (2).

Equation (1) ⇒

          $a+3(2)=7\\\\a+6=7\\\\a=1$

Therefore the sum of the n terms is given as,

             $S_{n}=\dfrac{n}{2}\left( 2(1)+\left( n-1\right) 2\right)\\\\\left=\dfrac{n}{2}\left( 2+\left2n-2\right)\\\\\\S_{n}=\dfrac{n}{2}\left( 2n)\\\\S_{n}=n^{2}$

Hence $n^{2}$ is the sum of the first n terms of this AP.