Math, asked by dhanushreddykalasani, 1 year ago

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

Answers

Answered by mysticd
8

 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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Answered by Anonymous
13

 \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.

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