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# Find the sum of the first 51 terms of an A.P whose second and third terms are 14 and 18 respectively.

Last updated date: 05th Mar 2024
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Hint:
Here we will use the formula of the general term of the AP to get the required value of the first term and the common difference. Then we will substitute the first term and the common difference in the formula of the sum of the AP to get the required answer.

Formula used:
We will use the following formulas:
1) The formula sum of the AP is ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ where,$a$ refers to the first term and $d$ refers to the common difference.
2) The formula for the general term of an AP is given by ${T_n} = a + \left( {n - 1} \right)d$.

Complete step by step solution:
Here we need to find the sum of the given AP. The number of terms of AP is equal to 51 i.e. $n = 51$.
Let the first term of the given AP be $a$ and the common difference of the given AP be $d$.
It is given that the second term of the AP is equal to 14.
So substituting $n = 2$ and ${T_n} = 14$ in the formula ${T_n} = a + \left( {n - 1} \right)d$, we get
$a + \left( {2 - 1} \right)d = 14$
On further simplifying the terms, we get
$\Rightarrow a + d = 14$ …………… $\left( 1 \right)$
It is given that the third term of the AP is equal to 18.
So substituting $n = 3$ and ${T_n} = 18$ in the formula ${T_n} = a + \left( {n - 1} \right)d$, we get
$a + \left( {3 - 1} \right)d = 18$
On further simplifying the terms, we get
$\Rightarrow a + 2d = 18$ …………… $\left( 2 \right)$
Now, we will subtract the equation $\left( 1 \right)$ from equation $\left( 2 \right)$, we get
$\begin{array}{l}a + 2d - \left( {a - d} \right) = 18 - 14\\ \Rightarrow a + 2d - a + d = 4\end{array}$
Adding and subtracting the like terms, we get
$\Rightarrow d = 4$
Now, we will substitute the value of the common difference, $d = 4$ in equation $\left( 1 \right)$.
$a + 4 = 14$
Now, we will subtract the number 4 from both sides.
$\begin{array}{l} \Rightarrow a + 4 - 4 = 14 - 4\\ \Rightarrow a = 10\end{array}$
Therefore, the first term of the AP is equal to 10.
Now, we will find the sum of the 51 terms of the AP.
Now, we will substitute the value of the first term, common difference and the number of terms in an AP in the formula ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$.
${S_{51}} = \dfrac{{51}}{2}\left( {2 \times 10 + \left( {51 - 1} \right) \times 4} \right)$
On further simplifying the terms, we get
$\Rightarrow {S_{51}} = \dfrac{{51}}{2}\left( {20 + 50 \times 4} \right)$
Multiplying the terms, we get
$\Rightarrow {S_{51}} = \dfrac{{51}}{2}\left( {20 + 200} \right)$
On adding the terms, we get
$\Rightarrow {S_{51}} = \dfrac{{51}}{2} \times 220$
On multiplying the numbers, we get
$\Rightarrow {S_{51}} = 5610$

Therefore, the required value of the sum of the AP is equal to 5610.

Note:
Here AP stands for Arithmetic Progression and it is defined as the sequence of the numbers which are in order and in which the difference of any two consecutive numbers is a constant value. A real life example of AP is when we add a fixed amount in our money bank every week. Similarly, when we ride a taxi, we pay an amount for the initial kilometer and pay a fixed amount for all the further kilometers, this also turns out to be an AP.