Question

In an A.P. of 10 terms, the sum of terms at odd positions is 25 and the sum of terms at even positions is 35. Find the value of $\sum_{1 \leq i < j \leq 10} a_i a_j$.

Answer 1

1455

Answer 2

Given $S_{odd} = 25$ and $S_{even} = 35$. The sum of all terms is $S_{10} = S_{odd} + S_{even} = 25 + 35 = 60$.

We know the identity: $(\sum_{i=1}^{n} a_i)^2 = \sum_{i=1}^{n} a_i^2 + 2 \sum_{1 \leq i < j \leq n} a_i a_j$

To find $\sum_{1 \leq i < j \leq 10} a_i a_j$, we need $\sum_{i=1}^{10} a_i^2$.

Let the A.P. be $a_1, a_2, \ldots, a_{10}$. $S_{odd} = a_1 + a_3 + a_5 + a_7 + a_9 = 25$ $S_{even} = a_2 + a_4 + a_6 + a_8 + a_{10} = 35$

Consider the differences between consecutive even and odd terms: $a_2 - a_1 = d$ $a_4 - a_3 = d$ ... $a_{10} - a_9 = d$

Summing these differences: $(a_2+a_4+...+a_{10}) - (a_1+a_3+...+a_9) = 5d$ $S_{even} - S_{odd} = 5d$ $35 - 25 = 5d$ $10 = 5d \implies d = 2$.

Also, $a_1 + a_3 + \dots + a_9 = 5a_5 = 25 \implies a_5 = 5$. And $a_2 + a_4 + \dots + a_{10} = 5a_6 = 35 \implies a_6 = 7$. This confirms $d = a_6 - a_5 = 7 - 5 = 2$.

Using $a_5 = a_1 + 4d$, we get $5 = a_1 + 4(2) \implies a_1 = 5 - 8 = -3$. The terms are $a_i = a_1 + (i-1)d = -3 + (i-1)2 = 2i - 5$.

Now, we calculate $\sum_{i=1}^{10} a_i^2$: $\sum_{i=1}^{10} (2i-5)^2 = \sum_{i=1}^{10} (4i^2 - 20i + 25)$ $= 4 \sum_{i=1}^{10} i^2 - 20 \sum_{i=1}^{10} i + \sum_{i=1}^{10} 25$ Using $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ and $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$: $= 4 \left(\frac{10(11)(21)}{6}\right) - 20 \left(\frac{10(11)}{2}\right) + 25(10)$ $= 4 (385) - 20 (55) + 250$ $= 1540 - 1100 + 250 = 690$.

Finally, using the identity: $2 \sum_{1 \leq i < j \leq 10} a_i a_j = (\sum_{i=1}^{10} a_i)^2 - \sum_{i=1}^{10} a_i^2$ $2 \sum_{1 \leq i < j \leq 10} a_i a_j = (60)^2 - 690$ $2 \sum_{1 \leq i < j \leq 10} a_i a_j = 3600 - 690$ $2 \sum_{1 \leq i < j \leq 10} a_i a_j = 2910$ $\sum_{1 \leq i < j \leq 10} a_i a_j = \frac{2910}{2} = 1455$.