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Sequences and series Quiz 2

mathematics

 Find the $10^{th}$ term of the AP $16, 11, 6, 1, - 4, ...?$

 Using first term and the common difference apply the formula ${T_n} = a + (n - 1)d$ to find the $10^{th}$ term of the AP.

In the given AP, we have $a = 16$ , $d = 11 - 16 = - 5$ and $n = 10$ . $ \\ {T_n} = a + (n - 1)d \\  \Rightarrow {T_{10}} = 16 + (10 - 1)( - 5) \\  \Rightarrow {T_{10}} = 16 + 9( - 5) \\ \Rightarrow 16 - 45 = - 29 \\ $ So, the $10^{th}$ term is -29.

 What term of the AP $11, 8, 5, 2, ....$ is -43?

 Using first term and the common difference apply the formula ${T_n} = a + (n - 1)d$ to find the $n$ .

 In the given AP, we have $a = 11$ , $d = (8 - 11) = - 3$ and ${T_n} = - 43$ . $ \\ \therefore {T_n} = a + (n - 1)d \\  \Rightarrow - 43 = 11 + (n - 1)( - 3) \\  \Rightarrow - 43 = 11 - 3n + 3 \\  \Rightarrow 3n = 11 + 3 + 43 \\  \Rightarrow n = \dfrac{{57}}{3} = 19 \\ $ Therefore, the $19^{th}$ term of the given AP is -43.

 Find the missing terms of the series $64, 61, 58, ?, 52$

 Find out the common difference of the series using $d = (a_{n + 1} – a_n )$ or $d = (a_n – a_{n-1})$.

 Given series has a common difference of $ - 3$ , i.e, $ \\ 64 - 3 = 61 \\ 61 - 3 = 58 \\ 58 - 3 = 55 \\ $ and, $55 - 3 = 52$

 Explore the given sequence $94, 91, 85, 79, 70$ and justify.

 Check the difference of every proceeding term.

In the given sequence- $94, 91, 85, 79, 70$ $ \\ $  The $3^{rd}$ term is incorrect, instead of 85, it should be 86. $ \\ $ First term $ = 94$ $ \\ $ Second term $ = 94 - 3 = 91$ $ \\ $ Third term $ = 91 - 5 = 86 \ne 85$ $ \\ $ Fourth term $ = 86 - 7 = 79$ $ \\ $ Last term $ = 79 - 9 = 70$ $ \\ $ Therefore, the given third term is incorrect.

 There are the following 4 sequences - $4, 7, 10, 13, .....$ ; $16, 21, 26, 31, ....$ ; $27, 34, 41, 48, ....$ and $8, 17, 26, 35, ....$. What are the similarities or consequences between them?

 Examine the common differences of each sequence one by one, compare the common differences, and determine whether they are in pattern or not.

In $1^{st}$ sequence- $4, 7, 10, 13, .....$ $ \\ $ Common difference is $ = 10 - 7 = 7 - 4 = 3$ ; $ \\ $ In $2^{nd}$ sequence- $16, 21, 26, 31, ....$ $ \\ $ Common difference $ = 26 - 21 = 21 - 16 = 5$ ; $ \\ $ $ \\ $ In $3^{rd}$ sequence- $27, 34, 41, 48, ....$ $ \\ $ Common difference $ = 41 - 34 = 34 - 27 = 7$ and $ \\ $ In 4th sequence- $8, 17, 26, 35, ....$ $ \\ $ Common difference $ = 26 - 17 = 17 - 8 = 9$ . $ \\ $ In each of the four given sequences, the common differences are exceeding by 2 every time. $ \\ $ As a result, the common differences follow a pattern of $3, 5, 7, 9, $