Question

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

25th term?​

Answer 1

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Step-by-step explanation:

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

Now, first term,a= 3

$common \: difference, \: d = a2 - a = 14 - 3= 11$

$Let \: nth \: term \: of \: the \: AP \: is \: 99 \: more \: than \: 25th \: term.$

$So, an = 99+ a25$

$⇒ a + (n-1)d = 99 + a + 24d$

$⇒3+ (n - 1)11 = 99 + 3 + 24 \times 11$

$⇒3 + 11n - 11 = 99 + 3 + 264$

$⇒11n - 8 = 366$

$⇒11n = 374$

$⇒n = \frac{374}{11} = 34$

$hence \: 34 ^{th} \: term \: of \: AP \: is \: 99 \: more \: than \: {25}^{th} \: term$

$\pink{i \: hope \: it \: helpfull \: for \: you}$