Question

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Which term of the progression 4, 9, 14, 19, … is 109?



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Answer 1

Given :

• AP = 4, 9, 14, 19, ...

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To Find :

• Which term is 109

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Solution :

Let 109 be the nth term of AP 4, 9, 14, 19,...

$\tt : \implies a_n = 109$

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Now, we have to find which term is 109, n = ?

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Now, in AP 4, 9, 14, 19,...

• First term, a = 4
• Common difference, d = 9 - 4 = 5

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Now, we know that :

$\large \underline{\boxed{\tt{a_n = a + (n-1)d}}}$

$\tt : \implies 109 = 4 + (n-1)5$

$\tt : \implies 109 = 4 + 5n-5$

$\tt : \implies 109 = -1 + 5n$

$\tt : \implies 109 + 1= 5n$

$\tt : \implies 110 = 5n$

$\tt : \implies \cancel{\dfrac{110}{5}} = n$

$\tt : \implies 22 = n$

$\tt : \implies n = 22$

$\Large \underline{\boxed{\tt{n = 22}}}$

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Hence, 22nd term of AP 4, 9, 14, 19,... is 109.

Answer 2

Step-by-step explanation:

Hence the nth term of an AP is 22

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