Math, asked by srannliyasms, 5 hours ago

first term=5&nth term is 45.sum is 400,find the no. of terms and the d​

Answers

Answered by Anonymous
76

  \small\bold{first \:  term  \: a=5}  \\  \bold{last \:  term  \: l=45  \: and  \: S_n = 400}

 \bold{Since, S_n =  \frac{n}{2} (a + 1)}

 \bold{ \implies400 =  \frac{n}{2} (5 + 45)}

 \bold{ \implies400 =  \frac{n}{2} (50)}

  \bold{\implies400 = n(25)}

 \bold{ \implies \: n = 16}

Since, l=a+(n−1)d

⇒45=5+(16−1)d

⇒45−5=(15)d

⇒40=15d

 \implies \: d =  \frac{40}{15}

⇒d =  \frac{8}{3}

Answered by ItzAshi
548

Step-by-step explanation:

{\Large{\bf{\underline{\purple{Question :}}}}} \\

First term of AP is 5 and nth term is 45. Sum is 400, find the no. of terms and the d

{\Large{\bf{\underline{\purple{Solution :}}}}} \\

Given :

  • First term (a) = 5
  • Last term (l) = 45
  • Sum = 400

Since the last term is given, we can use the formula

\\ {\large{\red{✠}}} \:  \:  \:  \:  \: {\large{\boxed{\sf{S_n  \: = \:  \frac{n}{2}  \: (a  \: + \:  l)}}}} \\  \\

Putting values in formula

{\bold{\sf{⟼ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  400  \: = \:  \frac{n}{2} (5  \: +  \: 45)}}} \\  \\

{\bold{\sf{⟼ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  400  \: = \:  \frac{n}{2} × 50}}} \\  \\

{\bold{\sf{⟼ \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \frac{400 \:  × \:  2}{50}  \: =  \: n}}} \\  \\

{\bold{\sf{⟼ \:  \:  \:  \:  \:  \:  \:  \:  \:  \: 8 \:  \times  \: 2 \:  =  \: n}}} \\  \\

{\bold{\sf{⟼ \:  \:  \:  \:  \:  \:  \:  \:  \:  \: 16  \: =  \: n}}} \\  \\

{\large{\sf{\fbox{\red{So, number of terms \:  = \:  16}}}}} \\  \\

Finding d,

For d we use the formula

{\large{\red{✠}}} \:  \:  \:  \:  \: {\large{\boxed{\sf{S_n  \: = \:  \frac{n}{2} \: [(2a  \: + \:  n  \: -  \: 1) d]}}}} \\  \\

{\large{\mathrm{\green{Putting   \:  \: \: S_n  \: = \:  400,  \:  \: n  \: =  \: 16,  \:  \: a  \: = \:  5}}}} \\  \\

{\large{\purple{⟹ \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }}}{\bold{\sf{400 \:  = \:  \frac{16}{2} \: [(2  \: × \:  5  \: + \:  16  \: -  \: 1)  \: × \:  d]}}} \\  \\

{\large{\purple{⟹ \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }}}{\bold{\sf{400 \:  = \:  8  \: (10  \: +  \: 15d)}}} \\  \\

{\large{\purple{⟹ \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }}}{\bold{\sf{\frac{400}{8} \: =  \: 10  \: + \:  15d}}} \\  \\

{\large{\purple{⟹ \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }}}{\bold{\sf{50 \: =  \: 10 \:  +  \: 15d}}} \\  \\

{\large{\purple{⟹ \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }}}{\bold{\sf{50 \: -  \: 10  \: = \: 15d}}} \\  \\

{\large{\purple{⟹ \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }}}{\bold{\sf{40  \: = \: 15d}}} \\  \\

{\large{\purple{⟹ \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }}}{\bold{\sf{\frac{40}{15}  \: = \:d}}} \\  \\

{\large{\purple{⟹ \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }}}{\bold{\sf{\frac{8}{3}  \: = \:d}}} \\  \\

{\large{\boxed{\mathrm{\red{So, common  \: difference \:  = \:  \frac{8}{3}}}}}} \\

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