The ninth term of an A.P. is equal to the seventh times the second term and the twelfth term exceeds five times the third term by

2 thoughts on “The ninth term of an A.P. is equal to the seventh times the second term and the twelfth term exceeds five times the third term by”

1. Answer :

   • First term, a = 1
   • Common difference, d = 6

   Solution :

   Given :

   • The ninth term of an A.P. is equal to the seven times the second term.
   • The twelfth term exceeds five times the third term by 2.

   To Find :

   • First term (a).
   • Common difference (d).

   Let’s solve,

   It is given that ,

   • 9th term is 7 times the second term.

   So,

   [tex]\:\:\:\:\: \red\bigstar\sf T_9 = 7\times T_2 [/tex]

   It can be also written as :

   [tex]\:\:\:\:\: \bf a+8d = 7(a+d) ……(i) [/tex]

   Solving (i) ,

   [tex]\implies \sf a + 8d = 7(a+d)\\ \sf \implies a + 8d = 7a + 7d \\\sf \implies 8d – 7d = 7a – a \\\implies\sf \pink{d=6a} ……(ii)[/tex]

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   • 12th term exceeds 5 times the 3rd term by 2.

   So,

   [tex]\:\:\:\:\: \red\bigstar\sf T_{12} = 5\times T_3 + 2 [/tex]

   It can be also written as :

   [tex]\:\:\:\:\: \bf a+11d = 5(a+2d) + 2……(iii) [/tex]

   Solving (iii) ,

   [tex]\implies\sf a+11d = 5(a+2d) + 2 \\\implies\sf a + 11d = 5a+10d+2 \\\implies\sf 11d-10d = 5a-a+2 \\\implies\sf \pink{d = 4a+2 }……(iv) [/tex]

   [tex]\sf \large\color{blue} \mathfrak{ Substituting\: (ii)\: in\: (iv) \:}[/tex]

   • d=6a ……(ii)

   • d = 4a+2 ……(iv)

   [tex]\implies \sf d = 4a+2 \\\implies\sf 6a = 4a+2 \\\implies\sf 2a= 2 \\\implies\sf \underline{\boxed{\mathfrak{\pink{a=1}}}}[/tex]

   [tex]\sf \large\color{blue} \mathfrak{ Putting\:the \:vaule\:of\: a\: in\: (iii) \:}[/tex]

   • d = 6a

   [tex]\implies\sf d=6(1) \\\implies\underline{\boxed{\mathfrak{\pink{d=6}}}}[/tex]

   Hence,

   First term(a) is 1 and Common difference(d) is 6.

   [tex]{\fcolorbox{red}{blue}{\orange{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: SugarCrash\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}} [/tex]

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2. Answer:

   Let the first term of A.P. be ‘a’ and the common difference be ‘d’

   Given,

   a₉ = 7a₂

   or, a+8d = 7(a+d) ________(i)

   and

   a₁₂ = 5a₃ + 2

   or, a+11d = 5(a+2d) + 2 ________(ii)

   from eq. (i)

   a+8d = 7a+7d

   -6a + d = 0 ________(iii)

   from eq. (ii)

   a+11d = 5a+10d + 2

   -4a + d = 0 ________(iv)

   Substracting eq. (iv) from eq. (iii), we get,

   -2a = 2

   a = 1

   or,

   from eq. (iii)

   -6a + d = 0

   d = 6

   Hence,

   First term is 1

   Common difference is 6

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