Question

Find the number of terms of an arithmetic progression whose third term is 20, 8th term is 10 and the sum of terms is 144. Explain the double answer.

Answer 1

Answer:

$\huge{\red{\boxed{\boxed{\green{\sf{n=9\:and\:16}}}}}}$

Step-by-step explanation:

$\huge{\red{\boxed{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}}$

□a3=20

□a8=10

□Sn=144

To find:

○n=?

$a3 = 20 \\ a + 2d = 20 - - - - - (1) \\ \\ a8 = 10 \\ a + 7d = 10 - - - - - (2) \\ \\ subtracting \:( 2) \: from \:( 1) \\ = ) a + 2d - (a + 7d) = 20 - 10 \\ = )a + 2d - a - 7d = 10 \\ = ) - 5d = 10 \\ = )d = \frac{ - 10}{5} \\ = )d = - 2 \\ \\ putting \: value \: of \: d \: in \: (1) \\ = ) a + 2d = 20 \\ = )a + 2 \times - 2 = 20 \\ = )a = 20 + 4 \\ = )a = 24 \\ \\ sn = \frac{n}{2} (2a + (n - 1)d) \\ = )144 = \frac{n}{2} (2 \times 24 + (n - 1) - 2) \\ = )144 = \frac{n}{2} (48 - 2n + 2) \\ = )144 = \frac{n}{2} \times 2(24 - n + 1) \\ = )144 = 24n - {n}^{2} +n \\ = ) {n}^{2} - 25n + 144 = 0 \\ = ) {n}^{2} - 16n - 9n + 144 = 0 \\ = )n(n - 16) - 9(n - 16) = 0\\ = )(n - 9)(n - 16) = 0\\ = )n - 9 = 0 \\ = ) n = 9 \\ = )n - 16 = 0 \\ = )n = 16 \\ \\ > > when \: we \: take \: n = 9 \: and \: get \: there \\ sum \: its \: results \: 144 \\ > > and \: when \: we \: take \: n = 16 \: for \\ \: sum \: of \: 16th \: term \: its \: also \: give \: \\ 144 \: because \: common \: difference \: \\ is \: in \: negtive$

$\huge{\red{\boxed{\boxed{\green{\sf{n=9\:and\:16}}}}}}$

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Answer 2

Answer:

n=9,16

Step-by-step explanation:

hope this helps !! thank you