Given the first 5 terms, write down the general term (a) 80,74,68,62,56… (b)3,1,13,19,127,… (c) 45,−811,1617,−3223,6429,…

We are given the first terms and we have to find the general form 1)80,74,68,62,56… a2−a=a3−a2=a4−a3=a5−a4 74−80=687−74=62−68=56−62 −6=−6=−6=−6 Hence, we have an=a+(n−1)d 80+(n+1)(−6) 80−6n+6=86−6n - general term, where n is positive integers

(b)3,1,13,19,127,… Here d=13 Thus, the above terms are in AP With first term =3 and common ratio =13 We get 3(13)n−1=9(13)n is the general form (c) 45,−811,1617,−3223,6429,… The general form is (−1)n−1(2)n+16n−1 Where n is positive integers

We are given the first terms and we have to find the general form 1)80,74,68,62,56… a2−a=a3−a2=a4−a3=a5−a4 74−80=687−74=62−68=56−62 −6=−6=−6=−6 Hence, we have an=a+(n−1)d 80+(n+1)(−6) 80−6n+6=86−6n - general term, where n is positive integers

(b)3,1,13,19,127,… Here d=13 Thus, the above terms are in AP With first term =3 and common ratio =13 We get 3(13)n−1=9(13)n is the general form (c) 45,−811,1617,−3223,6429,… The general form is (−1)n−1(2)n+16n−1 Where n is positive integers

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Kayleigh Dixon

Answered question

2022-01-24

Given the first 5 terms, write down the general term
(a)

80, 74, 68, 62, 56 \dots

(b)

3, 1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots

(c)

\frac{4}{5}, - \frac{8}{11}, \frac{16}{17}, - \frac{32}{23}, \frac{64}{29}, \dots

Answer & Explanation

trovabile4p

Beginner2022-01-25Added 13 answers

We are given the first terms and we have to find the general form
1)

80, 74, 68, 62, 56 \dots

a_{2} - a = a_{3} - a_{2} = a_{4} - a_{3} = a_{5} - a_{4}

74 - 80 = 687 - 74 = 62 - 68 = 56 - 62

- 6 = - 6 = - 6 = - 6

Hence, we have

a_{n} = a + (n - 1) d

80 + (n + 1) (- 6)

80 - 6 n + 6 = 86 - 6 n

- general term, where n is positive integers

Palandriy0

Beginner2022-01-26Added 14 answers

(b)

3, 1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots

Here

d = \frac{1}{3}

Thus, the above terms are in AP
With first term

= 3

and common ratio

= \frac{1}{3}

We get

3 {(\frac{1}{3})}^{n - 1} = 9 {(\frac{1}{3})}^{n}

is the general form
(c)

\frac{4}{5}, - \frac{8}{11}, \frac{16}{17}, - \frac{32}{23}, \frac{64}{29}, \dots

The general form is

\frac{{(- 1)}^{n - 1} {(2)}^{n + 1}}{6 n - 1}

Where n is positive integers

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