How many terms are in the arithmetic sequence 5, 1, −3, . . . , −111?

1 answer:

6 0

Here, it helps to find a general expression for the nth term in the sequence. Since the sequence is arithmetic, the general term is a_n = a + bn, where b is the common difference. By inspection of the first few terms we see the common difference is -4. Then by solving a_1 = a - 4x1 = 5, we obtain a = 9, so the nth term is: a_n = 9-4n. We then solve a_n=-111 for n, so 9-4n=-111. Rearranging, we get 4n=9+111, or 4n=120, or n=30. Thus there are 30 terms in the sequence.

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The sum of two numbers is 9 and the sum of their squares is 41. What is the larger number?

Sonbull [250]

The larger number is 5.

<u><em>Explanation</em></u>

Lets assume, the larger number is $a$ and the smaller number is $b$

As the sum of two numbers is 9, so...

$a+b= 9 ...............................(1)$

Now, the sum of their squares is 41, so....

$a^2 + b^2 = 41 .......................................(2)$

First, solving equation (1) for $b$ ....

$b=9-a$

Now, plugging this $b=9-a$ into equation (2) , we will get...

$a^2 +(9-a)^2 = 41\\ \\ a^2 +81-18a+a^2 =41 \\ \\ 2a^2-18a+81 =41 \\ \\ 2a^2-18a+40=0\\ \\ 2(a^2 -9a+20)=0\\ \\ a^2 -9a+20=0\\ \\ (a-5)(a-4)=0\\ \\ a=5,4$