Question

how many terms of ap 16 14,12 are needed to give the sum 60 explain why do we get two answers​

Answer 1

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Answer:-

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$\begin{gathered}\begin{gathered}\bf Given - \begin{cases} &\sf{an \: ap \: series \: 16, \: 14, \: 12, \: .\:. \: .} \\ &\sf{S_n \: = \: 60} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf To \: Find :- \begin{cases} &\sf{number \: of \: terms, \: n} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\blue{S_n\:=\dfrac{n}{2} \: ( \: 2\:a\:+\:(n\:-\:1)\:d \: )}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is sum of first n terms.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

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$\large\underline\purple{\bold{Solution :- }}$

$\begin{gathered}\begin{gathered}\bf Gɪᴠᴇɴ : \begin{cases} &\sf{First \: term, \: a \: = \: 16} \\ &\sf{Comon \: difference, d = 14 - 16 = - 2.}\\ &\sf{Sum \: of \: terms, \: S_n = 60}\\ &\sf{Let \: number \: of \: terms \: \: be \: n} \end{cases}\end{gathered}\end{gathered}$

$\sf \:  ⟼S_n \: = \: 60$

$\sf \:  ⟼ \:\dfrac{n}{2} \: ( \: 2\:a\:+\:(n\:-\:1)\:d \: ) \: = 60$

$\sf \:  ⟼\dfrac{n}{2} (2 \times 16 + (n - 1)( - 2) \: ) = 60$

$\sf \:  ⟼n \: (32 - 2n + 2) = 120$

$\sf \:  ⟼n \: (34 \: - \: 2n) \: = \: 120$

$\sf \:  ⟼ \: 34n \: - \: {2n}^{2} \: = \: 120$

$\sf \:  ⟼ {2n}^{2} \: - \: 34n \: + 120 = 0$

$\sf \:  ⟼ {n}^{2} - 17n \: + 60 \: = \: 0$

$\sf \:  ⟼ {n}^{2} - 12n - 5n + 60 = 0$

$\sf \:  ⟼n(n - 12) - 5(n - 12) = 0$

$\sf \:  ⟼(n - 12)(n - 5) = 0$

$\bf\implies \:n \: = \: 12 \: or \: n \: = \: 5$

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