Question

in an ap thrise the second term is equelent to eighth term and the sum of the fourth term and the seventh term is 9 grater then the ninth term find the AP soluti in steps​

Answer 1

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

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$\large{\green{\underline \bold{\tt{Given :-}}}}$

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• Thrice the second term is equal to eighth term

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• Sum of the fourth term and the seventh term is 9 greater then the ninth term

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$\large{\red{\underline \bold{\tt{To \: Find :-}}}}$

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• The AP

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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$\underline{\bold{\texttt{We know that :}}}$

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$\sf a_n = a + (n - 1)d$ ------ (1)

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• $\sf a_n$ = nth term

• a = First term

• n = Number of term

• d = Common difference

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$\underline{\bold{\texttt{Second term :}}}$

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• n = 2

• $\sf a_n = a_2$

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Putting these values in (1)

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➜ $\sf a_n = a + (n - 1)d$

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➜ $\sf a_2 = a + (2 - 1)d$

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➜ $\sf a_2 = a + d$ -------- (2)

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$\underline{\bold{\texttt{Eight term :}}}$

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• n = 8

• $\sf a_n = a_8$

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Putting these values in (1)

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➜ $\sf a_n = a + (n - 1)d$

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➜ $\sf a_8 = a + (8 - 1)d$

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➜ $\sf a_8 = a + 7d$ -------- (3)

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$\underline{\bold{\texttt{Fourth term :}}}$

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• n = 4

• $\sf a_n = a_4$

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Putting these values in (1)

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➜ $\sf a_n = a + (n - 1)d$

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➜ $\sf a_4 = a + (4 - 1)d$

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➜ $\sf a_4 = a + 3d$ -------- (4)

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$\underline{\bold{\texttt{Seventh term :}}}$

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• n = 7

• $\sf a_n = a_7$

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Putting these values in (1)

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➜ $\sf a_n = a + (n - 1)d$

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➜ $\sf a_7 = a + (7 - 1)d$

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➜ $\sf a_7 = a + 6d$ -------- (5)

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$\underline{\bold{\texttt{Ninth term :}}}$

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• n = 9

• $\sf a_n = a_9$

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Putting these values in (1)

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➜ $\sf a_n = a + (n - 1)d$

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➜ $\sf a_9 = a + (9 - 1)d$

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➜ $\sf a_9 = a + 8d$ -------- (6)

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$\purple{\underline \bold{According \: to \: the \ question :}}$

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Case I

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〚 Thrice the second term is equal to eighth term 〛

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$\pink{\bold{3 × Equation (2) = Equation (3)}}$

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➜ 3 × (a + d) = a + 7d

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➜ 3a + 3d = a + 7d

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➜ 3a - a = 7d - 3d

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➜ 2a = 4d

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➜ a = 2d -------- (7)

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Case II

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〚 Sum of the fourth term and the seventh term is 9 greater then the ninth term 〛

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$\pink{\bold{Equation (4)+Equation (5)=Equation (6) + 9}}$

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➜ a + 3d + a + 6d = a + 8d + 9

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➜ 2a + 9d = a + 8d + 9

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➜ 2a - a + 9d - 8d = 9

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➜ a + d = 9 --------- (8)

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$\underline{\bold{\texttt{Putting a = 2d from (7) to (8) }}}$

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➜ a + d = 9

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➜ 2d + d = 9

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➜ 3d = 9

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➜ $\sf d = \dfrac { 9 } { 3 }$

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➨ d = 3

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• Hence common difference is 3

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$\underline{\bold{\texttt{Putting d = 3 in (7) }}}$

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➠ a = 2d

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➜ a = 2(3)

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➨ a = 6

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• Hence the first term is 6

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So the AP will be :

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• a + 0d = 6 + 0(3) = 6

• a + 1d = 6 + 1(3) = 9

• a + 2d = 6 + 2(3) = 12

• a + 3d = 6 + 3(3) = 15

• a + 4d = 6 + 4(3) = 18

• a + 5d = 6 + 5(3) = 21

• a + 6d = 6 + 6(3) = 24

• a + 7d = 6 + 7(3) = 27

• a + 8d = 6 + 8(3) = 30

• a + 9d = 6 + 9(3) = 33

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Hence the AP will be 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 , 30 , 33 ..........