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1. Give the arithmetic sequence of 5 terms if the first term is 8 and the last term is 100.

2. Find the 9th term of the arithmetic sequence with a_1 = 10 and d = 1/2

3. Find a_1, if a_8 = 54 and a_9 = 60​

User Bayda
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1 Answer

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{ \qquad\qquad\huge\underline{{\sf Answer}}}

Here we go ~

Problem 1 :

First term (
{ \sf a_1})= 8

Number of terms (n) : 5

Last term (
{ \sf a_((last)) }) = 100

Now, the last term can be represented as :


\qquad \sf  \dashrightarrow \: a_((last)) = a_1 + (n - 1)d

[ n = 5, as it has 5 terms and d is common difference ]


\qquad \sf  \dashrightarrow \: 100 = 8 + (5- 1)d


\qquad \sf  \dashrightarrow \: 100 = 8 + 4d


\qquad \sf  \dashrightarrow \: 4d = 100 - 8


\qquad \sf  \dashrightarrow \: d = 92 / 4


\qquad \sf  \dashrightarrow \: d = 23

So, the terms of the sequence differ by 23

Therefore, the required sequence is :

  • 8, (8 + 23), (8 + 23 + 23), (8 +23 + 23 + 23), 100

  • 8, 31, 54, 77, 100

Peoblem 2 :

First term (
{ \sf a_1}) : 10

Common difference (d) : 1/2

9th term (
{ \sf a_((9))}) = ??

The general formula for a nth term in an Arithmetic sequence is :


\qquad \sf  \dashrightarrow \: \sf a_n = a_1 + (n - 1)d

[ n = 9, as we have to find 9th term of sequence ]


\qquad \sf  \dashrightarrow \: \sf a_((9)) = 10 + (9 - 1) \sdot (1)/(2)


\qquad \sf  \dashrightarrow \: \sf a_((9)) = 10 + (8) \sdot (1)/(2)


\qquad \sf  \dashrightarrow \: \sf a_((9)) = 10 + 4


\qquad \sf  \dashrightarrow \: \sf a_((9)) = 14

So, the 9th term of the sequence is 14

Problem 3 :

The 8th term and 9th term differ by : 60 - 54 = 6

so, the common difference (d) = 6

8th term can be represented as :


\qquad \sf  \dashrightarrow \: \sf a_((8))= a_1 + (8 - 1)d


\qquad \sf  \dashrightarrow \: \sf a_((8))= a_1 + 7d

Let plug the value of 8th term, and common difference ~


\qquad \sf  \dashrightarrow \: \sf a_1 + 7(6)= 54


\qquad \sf  \dashrightarrow \: \sf a_1 + 42= 54


\qquad \sf  \dashrightarrow \: \sf a_1= 54 - 42


\qquad \sf  \dashrightarrow \: \sf a_1= 12

Hence, 1st Term of the sequence is 12

User Jean Monet
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