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The n-th term of an AP is ${T_n} = 5n - 3$. Find the sum of first $20$ terms of an AP.

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Last updated date: 13th Jun 2024
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Answer
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Hint: By substituting the value of n, different terms can be found of an Arithmetic progression.
The sum of n terms of an AP is, ${S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]$
Where, $a = $ first term of an AP,${a_n} = $ last term of an AP and $n = $ number of terms whose sum is to be determined.

Complete step-by-step answer:
Given information:
The n-th term of an arithmetic progression is given as,
${T_n} = 5n - 3 \cdots \left( 1 \right)$
Substituting different values of n, the terms of an arithmetic progression can be generated.
Put $n = 1$ in equation (1), we get the first term as,
$
\Rightarrow {T_1} = 5\left( 1 \right) - 3 \\
\Rightarrow {T_1} = 2 \\
 $
Put $n = 2$ in equation (1), we get the second term as,
$
\Rightarrow {T_2} = 5\left( 2 \right) - 3 \\
\Rightarrow {T_2} = 7 \\
 $
Put $n = 3$ in equation (1), we get the third term as,
$
\Rightarrow {T_3} = 5\left( 3 \right) - 3 \\
\Rightarrow {T_3} = 12 \\
 $
Put $n = 4$ in equation (1), we get the fourth term as,
$
\Rightarrow {T_4} = 5\left( 4 \right) - 3 \\
\Rightarrow {T_4} = 17 \\
 $
The common difference of an arithmetic progression is calculated by subtracting previous term from the next term,
$\Rightarrow d = {T_2} - {T_1} \cdots \left( 2 \right)$
Substitute the value of ${T_2}$ and ${T_1}$ in equation (2), we get
$
\Rightarrow d = 7 - 2 \\
\Rightarrow d = 5 \\
 $
To obtain the last term or 20-th term, Put $n = 20$ in equation (1).
$
\Rightarrow {T_{20}} = 5\left( {20} \right) - 3 \\
\Rightarrow {T_{20}} = 97 \\
 $
The sum n terms of an arithmetic progression is given by,
$\Rightarrow {S_n} = \dfrac{n}{2}\left[ {{T_1} + {T_n}} \right] \cdots \left( 3 \right)$
Substitute the value of$n = 20$, ${T_1} = 2$ and ${T_n} = 97$ in equation (3), we get
$
\Rightarrow {S_n} = \dfrac{{20}}{2}\left[ {2 + 97} \right] \\
\Rightarrow {S_n} = 10\left[ {99} \right] \\
\Rightarrow {S_n} = 990 \\
 $
Hence, the sum of 20 terms of an arithmetic progression whose n-th term is ${T_n} = 5n - 3$ is $990$ .

Note: The sequence is called as an arithmetic progression when the successive terms differ by a common difference .
The above Sum can also be calculated from another formula which is in terms of common difference as,
$\Rightarrow {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cdots \left( i \right)$
Substitute the value of$n$, ${T_1}$ or $a$ , $d$ in equation (i), we get
$
\Rightarrow {S_n} = \dfrac{{20}}{2}\left[ {2\left( 2 \right) + \left( {20 - 1} \right)5} \right] \\
\Rightarrow {S_n} = 10\left[ {4 + 95} \right] \\
\Rightarrow {S_n} = 990 \\
 $