<em>The </em><em>answer </em><em>of </em><em>quest</em><em>ion</em><em> </em><em>no.1</em><em> </em><em> </em><em>and </em><em>2</em><em> </em><em>is </em><em>1.</em>
<em>Well</em><em>,</em><em>the</em><em> </em><em>quest</em><em>ion</em><em> </em><em>of</em><em> </em><em>1</em><em> </em><em>and </em><em>2</em><em> </em><em>are</em><em> </em><em>same</em><em>.</em>
<em>Look </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em>
<em>The</em><em> </em><em>answer</em><em> </em><em>of</em><em> </em><em>question</em><em> </em><em>n</em><em>o</em><em> </em><em>3</em><em> </em><em>is</em><em> </em><em>7</em><em>.</em>
<em>Hope </em><em>it</em><em> </em><em>helps</em><em>.</em><em>.</em><em>.</em>
<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>
Answer:
Is that on YT
Step-by-step explanation:
Might have to experiment a bit to choose the right answer.
In A, the first term is 456 and the common difference is 10. Each time we have a new term, the next one is the same except that 10 is added.
Suppose n were 1000. Then we'd have 456 + (1000)(10) = 10456
In B, the first term is 5 and the common ratio is 3. From 5 we get 15 by mult. 5 by 3. Similarly, from 135 we get 405 by mult. 135 by 3. This is a geom. series with first term 5 and common ratio 3. a_n = a_0*(3)^(n-1).
So if n were to reach 1000, the 1000th term would be 5*3^999, which is a very large number, certainly more than the 10456 you'd reach in A, above.
Can you now examine C and D in the same manner, and then choose the greatest final value? Safe to continue using n = 1000.
2,4,6,8,10,0,0,0,0,0,1,4,9,16,25,1,2,3,4,5