Question

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

Answer 1

✏️ Development:

✍ 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.

`\large {\text {$ \sf 7 \longrightarrow Number \: Terms $}}`

`\large {\text {$ \sf x \longrightarrow rasion $}}`

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:

`\huge {\boxed {\blue {\sf t_(10), t_(11), {\bf t_(12)} = \{ -4 , -6 , \bf -8 \} } }}`

So t₁₂ is equal to -8.

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