Which term of the A.P. 3, 14, 25, 36, … will be 99 more than its

25th term?​

1 thought on “Which term of the A.P. 3, 14, 25, 36, … will be 99 more than its <br /><br />25th term?​”

1. [tex] \huge \mathfrak \red{Æñßwēr}[/tex]

   Step-by-step explanation:

   Let a be the first term and d be the common difference.

   Now, first term,a= 3

   [tex]common \: difference, \: d = a2 – a = 14 – 3= 11[/tex]

   [tex]Let \: nth \: term \: of \: the \: AP \: is \: 99 \: more \: than \: 25th \: term.[/tex]

   [tex]So, an = 99+ a25[/tex]

   [tex]⇒ a + (n-1)d = 99 + a + 24d[/tex]

   [tex]⇒3+ (n – 1)11 = 99 + 3 + 24 \times 11[/tex]

   [tex]⇒3 + 11n – 11 = 99 + 3 + 264[/tex]

   [tex]⇒11n – 8 = 366[/tex]

   [tex]⇒11n = 374[/tex]

   [tex]⇒n = \frac{374}{11} = 34[/tex]

   [tex]hence \: 34 ^{th} \: term \: of \: AP \: is \: 99 \: more \: than \: {25}^{th} \: term[/tex]

   [tex] \pink{i \: hope \: it \: helpfull \: for \: you}[/tex]

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