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Question: The \[{5^{th}}\] and \[{15^{th}}\] terms of an AP are \[13\] and \[ - 17\] respectively. Find the su...

The 5th{5^{th}} and 15th{15^{th}} terms of an AP are 1313 and −17 - 17 respectively. Find the sum of the first 2121 terms of the AP.

Explanation

Solution

In AP for nth{n^{th}} term there is a formula which is, nth{n^{th}} term or An = A + (n−1)dAn{\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{ }} = $$$[\dfrac{n}{2}][2a + (n - 1)d]$ where, Sn{\text{ }} = {\text{ }}sum{\text{ }}of{\text{ }}n{\text{ }}terms{\text{ }}of{\text{ }}AP.Wewillgetthevalueofthesumoffirst We will get the value of the sum of first21$$ terms.

Complete step-by-step solution
Step 1: We have been given 5th{5^{th}}term which is 13$$$$, Now on putting the values in An{\text{ }} = {\text{ }}A{\text{ }} + {\text{ }}\left( {n - 1} \right)d$$$$, we get
A5= A + 4d = 13 …….. eq. (1){A_5} = {\text{ }}A{\text{ }} + {\text{ }}4d{\text{ }} = {\text{ }}13{\text{ }} \ldots \ldots ..{\text{ }}eq.{\text{ }}\left( 1 \right)
Similarly, We have been given 15th{15^{th}}term which is - 17$$$$, Now on putting the values in An{\text{ }} = {\text{ }}A{\text{ }} + {\text{ }}\left( {n - 1} \right)d$$$$, we get
A15= A + 14d = −17    ………eq. (2){A_{15}} = {\text{ }}A{\text{ }} + {\text{ }}14d{\text{ }} = {\text{ }} - 17\;\; \ldots \ldots \ldots eq.{\text{ }}\left( 2 \right)
On subtracting the eq. (2)eq.{\text{ }}\left( 2 \right)from eq. (1),eq.{\text{ }}\left( 1 \right),we get

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.