If
the third term of an Ap is 7 and
9th derm is 19, find the 27th term
​

2 thoughts on “If<br />the third term of an Ap is 7 and<br />9th derm is 19, find the 27th term<br />​”

1. Answer:

   9ap is the correct answer

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2. [tex]\boxed {\underline {\mathbb {FINAL\:ANSWER:-}}}[/tex]

   [tex]\boxed{27^{th}\:term\:of\:ap=55}[/tex]

   [tex]\boxed {\underline {\mathbb {GIVEN:-}}}[/tex]

   • the 3rd term of an Ap is 7

   • the 9th term of an Ap is 19

   [tex]\boxed {\underline {\mathbb {TO\:FIND:-}}}[/tex]

   27th term of ap

   [tex]\boxed {\underline {\mathbb {FORMULA\:USED:-}}}[/tex]

   an=a+(n-1)d

   solving Ap using substitution method

   [tex]\boxed {\underline {\mathbb {THINGS\:TO\:ASSUME:-}}}[/tex]

   • Let the first term of ap as a

   • Let the common difference as d

   [tex]\boxed {\underline {\mathbb {SOLUTION:-}}}[/tex]

   So 3rd term of ap will be [tex]\implies 7=a+ 2d[/tex]

   And 9th term is 19 [tex]\implies 19=a+8d[/tex]

   Therefore let [tex]\boxed{7=a+ 2d}[/tex] as eq1

   [tex]\boxed{19=a+18d}[/tex] as eq2

   [tex]\boxed{a=7-2d}[/tex] as eq3

   Now put eq3 in eq2 we get-

   [tex]19=(7-2d)+8d[/tex] [← putted eq3 in eq2]

   [tex]19=7-2d+8d\\19-7=8d-2d\\12=6d\\d=\frac{12}{6}\\\boxed {d=2}[/tex]

   as we got d s let’s put in eq3 in order to get a

   [tex]a=7-2(2)[/tex] [← putted 2 in eq3]

   [tex]a=7-4[/tex]

   [tex]\boxed {a=3}[/tex]

   [tex]a27=a+(n-1)d\\[/tex]

   [tex]=3+(27-1)2[/tex] [← putted value in equation]

   [tex]=3+26 \times 2\\=3+52\\\boxed {a27=55}[/tex]

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