Question

The \[{5^{th}}\] and \[{15^{th}}\] terms of an AP are \[13\] and \[ - 17\] respectively. Find the sum of the first $21$ terms of the AP.

Answer 1

Hint: In AP for \[{n^{th}}\] term there is a formula which is, \[{n^{th}}\] term or \[An{\text{ }} = {\text{ }}A{\text{ }} + {\text{ }}\left( {n - 1} \right)d\]
where, A $ = $ first term of AP
n $ = $ number of terms in AP
d $ = $ difference between two consecutive terms in AP.
Using the above formula, we will obtain the value of A, n and d after solving, then by using the formula:
\[Sn{\text{ }} = \]$[\dfrac{n}{2}][2a + (n - 1)d]$ where, \[Sn{\text{ }} = {\text{ }}sum{\text{ }}of{\text{ }}n{\text{ }}terms{\text{ }}of{\text{ }}AP.\]
We will get the value of the sum of first \[21\] terms.

Complete step-by-step solution
Step 1: We have been given \[{5^{th}}\]term which is \[13\]\[,\] Now on putting the values in \[An{\text{ }} = {\text{ }}A{\text{ }} + {\text{ }}\left( {n - 1} \right)d\]\[,\] we get
\[{A_5} = {\text{ }}A{\text{ }} + {\text{ }}4d{\text{ }} = {\text{ }}13{\text{ }} \ldots \ldots ..{\text{ }}eq.{\text{ }}\left( 1 \right)\]
Similarly, We have been given \[{15^{th}}\]term which is \[ - 17\]\[,\] Now on putting the values in \[An{\text{ }} = {\text{ }}A{\text{ }} + {\text{ }}\left( {n - 1} \right)d\]\[,\] we get
\[{A_{15}} = {\text{ }}A{\text{ }} + {\text{ }}14d{\text{ }} = {\text{ }} - 17\;\; \ldots \ldots \ldots eq.{\text{ }}\left( 2 \right)\]
On subtracting the \[eq.{\text{ }}\left( 2 \right)\]from \[eq.{\text{ }}\left( 1 \right),\]we get
\[\begin{gathered}
  30{\text{ }} = {\text{ }} - 10{\text{ }}d \\
  d{\text{ }} = {\text{ }} - 3 \\
\end{gathered} \]
On putting the value of ‘d’ in \[eq.{\text{ }}\left( 1 \right),\]we get
\[\begin{array}{*{20}{l}}
  {13{\text{ }} = {\text{ }}A{\text{ }} + {\text{ }}4\left( { - 3} \right)} \\
  {A{\text{ }} = {\text{ }}25}
\end{array}\]
Step 2: Now to find sum of the first $21$ terms of the AP, we will use the formula\[,\] \[{S_{n{\text{ }}}} = \]$[\dfrac{n}{2}][2a + (n - 1)]$
\[{S_{21}} = \]$[\dfrac{{21}}{2}][2 \times 25 + (21 - 1)( - 3)]$
\[{S_{21}} = \] $[\dfrac{{21}}{2}][50 + 20$\[ \times ( - 3)]\]
\[{S_{21}} = \]$[\dfrac{{21}}{2}][50 - 60]$
\[{S_{21}} = \] $\dfrac{{21}}{2} \times ( - 10)$
\[{S_{21}} = \] $ - 105$
Therefore, sum of first $21$ terms of the AP is \[ - 105.\]

Note:Here, the sum is negative, which shows the value of negative terms is greater than positive terms in the given AP.