Question

in an ap.the sum of its first ten term is -80 and the sum of its next ten term is -280.find the ap​

Answer 1

Answer:

$\frac{ - 13}{5} \: \: \: \: \frac{ - 19}{5} \: \: \: - 5............$

Step-by-step explanation:

$s10 = - 80 \\ s20 = - 280 \\ by \: the \: formula \: sn = \frac{n}{2} (2a + (n - 1)d) \\ \frac{10}{2} (2a + 9d) = - 80 \\ 5(2a + 9d) = - 80 \\ 10a + 45d = - 80 \\ \div by \: 5 \\ 2a + 9d = - 16 - - - - - eq1 \\ \frac{20}{2} (2a + 19d) = - 280 \\ 10(2a + 19d) = - 280 \\ 20a + 190d = - 280 \\ \div by \: 10 \\ 2a + 19d = - 28 - - - - - - eq2 \\ 2a + 9d = - 16 \\ 2a + 19d = - 28 \\ 10d = - 12 \\ d = \frac{ - 6}{5} \\ 2a + 9( \frac{ - 6}{5} ) = - 16 \\ 2a + \frac{ - 54}{5} = - 16 \\ \frac{10a - 54}{5} = - 16 \\ 10a - 54 = - 80 \\ 10a = - 26 \\ a = \frac{ - 13}{5} \\ the \: ap \: is \: \frac{ - 13}{5} \: \: \: \frac{ - 19}{5} \: \: \: \: \: - 5.......$

Answer 2

Step-by-step explanation:

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let the first term and d is the common differences of AP

= Sn=n/2{2a+(n-1)d}

now.............

=S10=10/2{2a+(10-1)d}

= -80=5(2a+9d)

= 2a+9d=-16.........(1)

again....

S20-S10=-280

=20/2{2a+19d}+80=-280

= 10{2a+19d}=-360

= 2a+19d=-36.........(2)

solve this two equation

then..

we found

d=-2

a=1

so...

AP=1,-1,-3,-5......