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Find the twentieth term if a36 = 78 and a42 = 90.
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Hey are you there?
yes
are you given any other numbers?
like 1+2+3+..+..
that is all
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helloo?
sorry i didnt see your post
ok so \[a _{n}=a _{1}+\left( n-1 \right)d\]
\[78=a _{1}+\left( 36-1 \right)d\] \[78=a _{1}+35d\]
\[a _{1}=35d-78\]
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\[90=a _{1}+\left( 42-1 \right)d\] \[90=a _{1}+41d\] \[90=\left( 35d-78 \right)+41d\]
solve for d, do you understand?
168=96d 168/96=d
humm....
wow i dont know that is confusing
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\[a _{n}=a _{1}+(n-1)d\] \[a _{20}=a _{1}+(20-1)d\] \[a _{20}=a _{1}+19d\] \[d=168/96\]
is that final answer
are you sure thats all your problem says?
yes
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