Question

If the sum of the first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of the first n term.

Answer 1

We can use the formula of the sum of n terms in Arithmetic progression that is ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ where, a initial term of the AP and d is the common difference of successive numbers. Substitute the values n and ${S_7}$ after that calculate the value of a and d. Substitute the values a and d in ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$. Calculate the sum of the AP ${S_n}$.

Complete step by step answer:
Given data,
The sum of the first 7 terms of an AP is 49 and the sum of the 17 terms is 289.
We know about the formula of the sum of n terms in Arithmetic progression that is ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.
Now, calculate the value of $a$ and $d$. Substitute the values $n = 7$ and ${S_7} = 49$ in the expression ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.
$\Rightarrow 49 = \dfrac{7}{2}\left[ {2a + \left( {7 - 1} \right)d} \right]$
$\Rightarrow 14 = 2a + 6d$
$\Rightarrow 2a = 14 - 6d$
$\Rightarrow a = 7 - 3d$ ………………... (i)
We can calculate the value of $a$ and $d$. Substitute the values $n = 17$and ${S_7} = 289$ in the expression ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.
$\Rightarrow 289 = \dfrac{{17}}{2}\left[ {2a + \left( {17 - 1} \right)d} \right]$
$\Rightarrow 34 = 2a + 16d$
$\Rightarrow 2a = 34 - 16d$
$\Rightarrow a = 17 - 8d$ ……………….(ii)
Now, subtract the equation (i) from (ii) and obtain the value of d:
$\Rightarrow 17 - 8d = 7 - 3d$
$\Rightarrow 5d = 10$
$\Rightarrow d = 2$
Calculate the value of a by substituting the value of d in equation (i).
$\Rightarrow a = 7 - 3\left( 2 \right)$
$\Rightarrow a = 1$
Now, calculate the sum of the first n terms.
Now, calculate the value of ${S_n}$. Substitute the values $a = 1$ and $d = 2$ in the expression ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.
$\Rightarrow {S_n} = \dfrac{n}{2}\left[ {2\left( 1 \right) + \left( {n - 1} \right)2} \right]$
$= \dfrac{n}{2}\left[ {2n} \right]$
$= {n^2}$

Hence, the sum of the first n terms of an AP is ${S_n} = {n^2}$.

Note:
If we add or subtract the same numbers from the arithmetic progression series, then the resulting series are also in arithmetic progression with the same common difference.
If we divide or multiply the same non-zero numbers from the arithmetic progression series, then the resulting series are also in arithmetic progression with the same common difference.