Which term of the A.P. 3, 14, 25, 36, … will be 99 more than its 25th term? December 30, 2021 by Skylar Which term of the A.P. 3, 14, 25, 36, … will be 99 more than its 25th term?

[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] Log in to Reply

## [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]