If the {n^{th}} term of an A.P. be (2n - 1), | vedclass

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Question

(c) Given that ${T_n} = 2n - 1$

First term $a = 2\ .\ 1 - 1 = 1$

Second term $b = 2\;.\;2 - 1 = 3$

Third term $c = 2\;.\;3 - 1 = 5$

Theref

Vedclass · Accepted Answer

${n^2}$

Vedclass · Answer

(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} $

$= 2\Sigma n - \Sigma \;1 = n(n + 1) - n = {n^2}$

If the ${n^{th}}$ term of an $A.P.$ be $(2n - 1)$, then the sum of its first $n$ terms will be

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