Fine t12 term in an arithmetic progression having t3=10 and t10=-4

✍ Having as t₃ =10 and t₁₀ = -4.  We have to find the common ratio.

❑  If, in front of the sequence from 3 to 10, there are 7 numbers, that is, 7 terms, we have to find a pattern, what I used was this one.

          We see that in the calculation, the only number that fits is -2.

➤ Because if we stop to think about it, we have -2 of common ratio among these 7 terms is -14, so 10-14 = -4

So the hypothesis is correct!

➜ Being right as -2 and resuming the terms of 10, we have:

So t₁₂ is equal to -8.