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My question is related to Arithmetic progression. I would solve it myself but I need some help understanding the question.

Q.The second, 31st and the last terms of an A.P are $$ \frac{31}{4},\frac{1}{2} $$ and $$ \frac{-13}{2} $$ respectively. Find the number of terms.

How can I use the $$ Tn=a+(n-1)d $$ formula here, I need some explanation of this question.

-Thanks.

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The second term is $31/4 = a + d$, the $31-$th term is $1/2 = a + (31-1)d$, solve these two equations for $a$ and $d$. then use $-13/2 = a + (n-1)d$ to solve for $n$. –  Santosh Linkha Feb 18 '13 at 18:37
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2 Answers 2

$a+d=\frac{31}{4}$, $a+30d=\frac12$, and $a+(l-1)d=-\frac{13}{2}$. Find $l$.

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The algebraic approach described by Jacob Black is the right one. For fun, let us see how someone innocent of algebra might tackle things.

From the $2$nd term to the $31$-th (which takes $29$ steps), we go down by $\dfrac{31}{4}-\dfrac{1}{2}=\dfrac{31}{4}-\dfrac{2}{4}=\dfrac{29}{4}$.

So we go down by $\dfrac{1}{4}$ each time.

From the $31$-th term to the end, we went down by $\dfrac{1}{2}-\left(\dfrac{-13}{2}\right)=\dfrac{14}{2}=\dfrac{28}{4}$. So it took $28$ steps. Thus the total number of terms is $31+28=59$.

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