- Home
- Standard 11
- Mathematics
8. Sequences and Series
easy
If the ${n^{th}}$ term of an $A.P.$ be $(2n - 1)$, then the sum of its first $n$ terms will be
A
${n^2} - 1$
B
${(2n - 1)^2}$
C
${n^2}$
D
${n^2} + 1$
Solution
(c) Given that ${T_n} = 2n – 1$
First term $a = 2\ .\ 1 – 1 = 1$
Second term $b = 2\;.\;2 – 1 = 3$
Third term $c = 2\;.\;3 – 1 = 5$
Therefore sequence is $1,\;3,\;5,…….2n – 1$.
Now sum of the first $n$ terms is ${S_n} = \frac{n}{2}[a + l]$
$ = \frac{n}{2}[1 + 2n – 1] = \frac{n}{2}\;.\;2n = {n^2}$
Aliter : Since ${T_n} = 2n – 1$
$ \Rightarrow {S_n} = \Sigma {T_n} $
$= 2\Sigma n – \Sigma \;1 = n(n + 1) – n = {n^2}$
Standard 11
Mathematics
