Question

Find the 23rd term of the sequence: 1,4,5,8,9,12,13,16,17,........

Answer 1

$Given \: sequence \: 1,4,5,8,9,12,13,16,17,\cdot\cdot\cdot$

/* Observe the pattern ,we find two sequences as follows*/

$1,5,9,13,17,\cdot\cdot\cdot \: ---(1)$

$and \\4,8,12,16,\cdot\cdot\cdot\: ---(2)$

$23^{rd} \: term \: of \: the \: sequence \\= 12^{th} \: term \: of \: the \: First \: sequence$

$1,5,9,13,17,\cdot\cdot\cdot \: ,is \: an\: A.P$

$First \: term (a) = 1$

$Common \: difference (d) = a_{2} - a_{1} \\= 5 - 1 \\= 4$

$\boxed { \pink { n^{th} \:term (a_{n}) = a + (n-1)d }}$

$Here, n = 12$

$a_{12} = a + (12 - 1 )d \\= a + 11d \\= 1 + 11 \times 4\\= 1 + 44 \\ = 45$

Therefore.,

$\red { 23^{rd} \: term \: of \: the \: sequence }$

$\green { = 45 }$

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Answer 2

Answer:

45

Step-by-step explanation:

For all the odd terms: 1,5,9,13,17

You add 4 to obtain

For all the even terms: 4,8,12,16,20

You also add 4 to obtain

So, for the 23rd term, which is odd, you add 4 to 1 for 11 times (1 + 11*4 = 1 + 44 = 45)

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