An algebra problem by Thomas Jacob

Let us assume the number of terms in this progression to be $2n$ .

Then the number of odd and even terms are equal i.e. $n$

The sum of the odd terms and even terms is given as $24$ and $30$ respectively.

We know $S_{n}$ = $\displaystyle \frac{n}{2}$ [ $2a$ $+$ $(n-1)$ $d$ ]

For the odd sequence, let the first term be $a$ and common difference be $2d$ ,

$d$ being the common difference of the original A.P with $2n$ terms.

For the even sequence, the first term is equal to $a + d$ and the common difference is $2d$ as well.

Plugging these values into the $S_{n}$ formula for odd and even terms, we get two equations.

For the odd sequence, we get,

$\displaystyle \frac{n}{2}$ [ $2a$ $+$ $(n-1)$ $2d$ ] = $24$

This can be simplified to, $an$ $+$ $nd( n - 1)$

For the even sequence,

$\displaystyle \frac{n}{2}$ [ $2(a+d)$ $+$ $(n-1)$ $2d$ ] = $30$

This expression can be further simplified to,

$(a + d) n + n ( n - 1) d = 30$

Solving these two equations, we get, $n d = 6$

Finally, since we know the difference between the first and last terms, we can write another equation,

$a_{n}$ = $a + (2 n - 1) d$

or, $a + 10.5$ = $a + (2 n - 1) d$

or, $(2n-1)d$ = $10.5$ .

Plugging in the value of $nd$ we found earlier, we get $d = 1.5$

Now, plugging in this value of $d$ into $n d = 6$ , we get $n = 4$

Since we wanted the number of terms in our original AP, the answer is $2n$ = $A$ = $8$

Suppose that there are $2n$ terms in the AP. Let the common difference be $d$ .

If we consider the odd-indexed sequence with the even-indexed sequence termwise, the difference of each pair of terms is $d$ , hence the difference of these two sequences is $n \times d = 30 - 24 = 6$ .

We are given that the last term exceed the first term by 10.5, hence $a+ (2n-1)d - a = (2n-1) d = 10.5$ .

Solving these 2 equations by substituting $d = \frac{6}{n}$ , we obtain $n = 4, d = 1.5$ .

Hence, there are $2n = 8$ terms in the sequence.