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irina1246 [14]
3 years ago
7

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

Mathematics
1 answer:
antiseptic1488 [7]3 years ago
8 0
<h2>✏️ Development:</h2>

✍ 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.

<h2>⊱───────────⊰✯⊱────────⊰</h2>

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In ​ kite PQRS ​, TQ=3 cm and TP=4 cm. What is SP ? Enter your answer in the box.
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Kite is a quadrilateral in which two disjoint pairs of consecutive sides are congruent

Given: In kite PQRS

where PR and SQ are the diagonals of Kite respectively as shown in the figure given below.

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Let PR is the main diagonal and SQ is the cross diagonal of kite PQRS as shown in figure,  also let T is the intersection point of PQRS.

By Property of Kite, diagonal SQ bisects PR at perpendicular angle i.e, 90 degree.

i,e

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