Question

If the nth term of an A.P; is an = 3+2n then S24​

Answer 1

$Given \: n^{th} \:term (a_{n}) = 3 + 2n$

$i ) If \: n = 1 \:then \\First \:term (a) = 3 + 2\times 1 \\= 3 + 2 \\= 5$

$ii ) If \: n = 24 \:then \\a_{24}= 3 + 2\times 24\\= 3 + 48\\= 51$

$\boxed { \pink { Sum \:of \:n \: terms (S_{n}) = \frac{n}{2} (a+a_{n}) }}$

$\red{ Sum \: of \: 24 \: terms (S_{24}) } \\= \frac{24}{2}( 5 + 51) \\= 12 \times 56 \\= 672$

Therefore.,

$\red{ Sum \: of \: 24 \: terms (S_{24}) }\green { = 672 }$

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Answer 2

$\large\bf\underline{Given:-}$

• nth term of AP is an = 3 + 2n

$\large\bf\underline {To \: find:-}$

• sum of 24 terms.

$\huge\bf\underline{Solution:-}$

$\dag \: \boxed { \large\rm \:a_n = a+(n-1)d}$

$\rm \dashrightarrow \: a_n = 3 + 2n \: .......(i)$

Putting n = 1 in eq.(i)

$\rm \dashrightarrow \:a_1 = 3+ 2 \times 1 \\ \\ \rm \dashrightarrow \:a_1 = 3 + 2 \\ \\ \rm \dashrightarrow \:a_1 = 5$

putting n = 2 in eq.(i)

$\rm \dashrightarrow \:a_2 = 3 + 2 \times 2 \\ \\ \rm \dashrightarrow \:a_2 = 3 + 4 \\ \\ \rm \dashrightarrow \:a_2 = 7$

putting n = 3 in eq.(i)

$\rm \dashrightarrow \:a_3 = 3 + 2 \times 3 \\ \\ \rm \dashrightarrow \:a_3 = 3 + 6 \\ \\ \rm \dashrightarrow \:a_3 = 9$

So, we get an AP :- 5 , 7 ,9

In which ,

• first term (a) = 5
• common difference (d) = 7-5 = 2

$\boxed{ \large \dag \rm \: S_n = \frac{n}{2} [2a + (n - 1)d]}$

$\rm \rightarrowtail \: S_{24} = \frac{24}{2} [2 \times 5 + 23 \times 2] \\ \\ \rm \rightarrowtail \: S_{24} = 12(10 + 46) \\ \\ \rm \rightarrowtail \: S_{24} = 12(56) \\ \\ \rm \rightarrowtail \: S_{24} = 672$

So,

»★ Sum of 24 terms = 672.