Difference between revisions of "1973 AHSME Problems/Problem 26"

Latest revision as of 13:03, 20 February 2020

Problem

The number of terms in an A.P. (Arithmetic Progression) is even. The sum of the odd and even-numbered terms are 24 and 30, respectively. If the last term exceeds the first by 10.5, the number of terms in the A.P. is

$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 8$

Solution

Let $a$ be the first term, $n$ be the number of terms, and $d$ be the common difference. That means the last term is $a+r(n-1)$ .

We can write an equation on the difference between the last and first term based on the conditions. $a+r(n-1)-a =10.5$ $rn-r=10.5$ Also, half of the terms add up to $24$ while the other half of the terms add up to $30$ , so $24 + r\frac{n}{2} = 30$ $nr = 12$ Substituting the value back to a previous equation, $12-r=10.5$ $r=1.5$ Substituting to a previous equation again, $1.5n-1.5=10.5$ $n=8$ Thus, there are $\boxed{\textbf{(E)}\ 8}$ terms in the arithmetic sequence.

1973 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 25	Followed by Problem 27
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

@@ Line 28: / Line 28: @@
 ==See Also==
-{{AHSME 35p box|year=1973|num-b=25|num-a=27}}
+{{AHSME 30p box|year=1973|num-b=25|num-a=27}}
 [[Category:Intermediate Algebra Problems]]

Page

Toolbox

Search

Difference between revisions of "1973 AHSME Problems/Problem 26"

Latest revision as of 13:03, 20 February 2020

Problem

Solution

See Also