Question

Find the sum of first 24 terms of the A.P., ${a_1},{a_2},{a_3},.............$ if it is known that ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$ .

Answer 1

Here, we will find the sum of the first 24 terms which are in A.P. We will use the arithmetic progression to find the value of each term and then substituting the value of the terms in the given condition and find an equation. We will use the sum of the first number of terms in an A.P., and then by substituting the equation of Arithmetic progression we will find the sum of the first 24 terms which are in A.P.

Formula Used:
We will use the following formula:
1.Arithmetic Progression is given by the formula ${T_n} = a + \left( {n - 1} \right)d$
2.Sum of the first $n$ terms in an A.P. is given by the formula ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ where $a$ is the first term, $d$ is the common difference and $n$ is the number of terms

Complete step-by-step answer:
Let the first term of the Arithmetic Progression be $a$ , the common difference be $d$ .
Arithmetic Progression is given by the formula ${T_n} = a + \left( {n - 1} \right)d$
We are given that ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$ .
$\Rightarrow$ The first term is given by the formula ${T_1} = a + \left( {1 - 1} \right)d = a$
$\Rightarrow$ The fifth term is given by the formula ${T_5} = a + \left( {5 - 1} \right)d = a + 4d$
$\Rightarrow$ The tenth term is given by the formula ${T_{10}} = a + \left( {10 - 1} \right)d = a + 9d$
$\Rightarrow$ The fifteenth term is given by the formula ${T_{15}} = a + \left( {15 - 1} \right)d = a + 14d$
$\Rightarrow$ The twentieth term is given by the formula ${T_{20}} = a + \left( {20 - 1} \right)d = a + 19d$
$\Rightarrow$ The twenty fourth term is given by the formula ${T_{24}} = a + \left( {24 - 1} \right)d = a + 23d$
Substituting the first term, the fifth term, the tenth term, the fifteenth term, the twentieth term, the twenty fourth term in the equation ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$ , we get
$\Rightarrow {a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$
$\Rightarrow a + a + 4d + a + 9d + a + 14d + a + 19d + a + 23d = 225$
By adding all the terms, we get
$\Rightarrow 6a + 69d = 225$
Dividing by 3, we get
$\Rightarrow 2a + 23d = 75$ ……………………………………………………………………………………………………$\left( 1 \right)$
Sum of the first $n$ terms in an A.P. is given by the formula ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Now, we will find the sum of the first 24 terms in an A.P.
By, substituting $n = 24$ , we get
$\Rightarrow$ Sum of the first 24 terms${S_{24}} = \dfrac{{24}}{2}\left[ {2a + \left( {24 - 1} \right)d} \right]$
By simplifying the equation, we get
$\Rightarrow$ Sum of the first 24 terms${S_{24}} = 12\left[ {2a + 23d} \right]$
By substituting equation $\left( 1 \right)$ , we get
$\Rightarrow$ Sum of the first 24 terms${S_{24}} = 12 \times 75$
$\Rightarrow$ Sum of the first 24 terms${S_{24}} = 900$
Therefore, the sum of the first 24 terms is $900$.

Note: We know that Arithmetic progression is a sequence of numbers where the difference between the two consecutive numbers is a constant. Arithmetic series is the sum of the terms of the arithmetic progression. An arithmetic sequence of numbers is also defined as a sequence of numbers where each number is the sum of the preceding number and common difference (d) is a constant. If the same number is added or subtracted from each term of an A.P., then the resulting terms in the sequence are also in A.P. but with the same common difference. The sum of the first $n$ natural numbers is the same as the Arithmetic series of $n$ terms.