Question

If the sum of 1st 7 terms of an aritmetic sequence is 49 and that of 17 terms is 289.
(a) Find common difference.
(b) Find 20th term.​

Answer 1

7×7=49 & 17×17=289

Step-by-step explanation:

20×20=400

Answer 2

$\begin{gathered}\begin{gathered}\bf Given - \begin{cases} &\sf{sum \: of \: first \: 7 \: terms \: is \: 49} \\ &\sf{sum \: of \: first \: 17 \: terms \: is \: 289} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf To \: Find :- \begin{cases} &\sf{common \: difference} \\ &\sf{ {20}^{th} \: term \: of \: series} \end{cases}\end{gathered}\end{gathered}$

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

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

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

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

• Sₙ is the sum of first n terms.

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

$\large\underline{\bold{❥︎Step :- 1 }}$

$\begin{gathered}\begin{gathered}\bf Let = \begin{cases} &\sf{first \: term \: be \: a} \\ &\sf{common \: difference \: be \: d}\end{cases}\end{gathered}\end{gathered}$

$\bf \:  ⟼  ✬ \: S_7 \: = \: 49$

$\bf\implies \:\dfrac{ \cancel{7}}{2} (2a + (7 - 1)d) = \cancel{49} \: 7$

$\bf\implies \:2a + 6d = 14$

$\bf\implies \: \: a + 3d \: = 7 - - - - (1)$

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$\large\underline{\bold{❥︎Step :- 2 }}$

$\bf \:  ⟼ \:  ✬ \: S_{17} \: = 289$

$\bf\implies \:\dfrac{ \cancel{17}}{2} (2a \: + (17 - 1)d) = \cancel{289} \: 17$

$\bf\implies \:2a + 16d \: = 34$

$\bf\implies \:a \: + \: 8d = 17 - - - - (2)$

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$\large\underline{\bold{❥︎Step :- 3 }}$

 ✬ On Subtracting equation (1) from equation (2), we get

$\bf \:  \cancel{a} \: + 8d \: - \cancel{a} \: - 3d \: = \: 17 - 7$

$\bf\implies \:5d \: = \: 10$

$\bf\implies \:d \: = 2$

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$\large\underline{\bold{❥︎Step :- 4 }}$

 ✬ On substituting the value of 'd' in equation (1), we get

$\bf\implies \:a \: + 3 \times 2 = 7$

$\bf\implies \:a \: = \: 1$

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$\large\underline{\bold{❥︎Step :- 5 }}$

☆ To find 20th term of AP series,

$\bf \:  ⟼ a_n \: = a \: + \: (n - 1)d$

$\bf\implies \:a_{20} \: = 1 + 19 \times 2$

$\bf\implies \:a_{20} \: = \: 39$

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$\begin{gathered}\begin{gathered}\bf Hence \: \begin{cases} &\sf{common \: difference \: = 2} \\ &\sf{ \:a_{20} \: = \: 39} \end{cases}\end{gathered}\end{gathered}$

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