Question

in an A.P 12th term is 23 and the sum of first four term is 24 then find the sum of first 20 term​

Answer 1

Answer:

20th term=

Step-by-step explanation:

d=34/19

a=63/19

Answer 2

$Let \: a \: and \: d \: are \: first \:term \: and \\common \: differnce \: of \:an \: A.P$

$12^{th} \:term= 23 \:( given )$

$\implies a + 11d = 23 \: ---(1)$

$Sum \: of \: of \: first\: 4 \:terms = 24$

$\boxed {\pink {Sum \:of \:n \:terms (S_{n}) = \frac{n}{2}[2a+(n-1)d }}$

$\implies \frac{4}{2}[ 2a + 3d] = 24$

$\implies 2(2a+3d) = 24$

$\implies 2a+3d= 12 \: --(2)$

$Multiplying \: Equation \:(2) \: by \: 2 , \: and \\subtract \: equation \: (2), we \:get$

$\implies 2a+22d -(2a+3d) = 46 - 12$

$\implies 2a+22d - 2a-3d= 46 - 12$

$\implies 19d = 34$

$\implies\pink { d = \frac{34}{19}} \: --(3)$

$Put \: d = \frac{34}{19} \: in \: equation \: (1), we \:get$

$\implies a + 11 \times \frac{34}{19} = 23$

$\implies a + \frac{374}{19} = 23$

$\implies a = 23 - \frac{374}{19}$

$\implies a = \frac{437 - 374}{19}$

$\implies \blue {a = \frac{63}{19}}$

$Sum \: of \: first \:20 \: terms \\= \frac{20}{2}[ 2\times \big(\frac{63}{19} \big) + (20-1)\times \frac{34}{19} ]$

$= 10[ \frac{126}{19} + 19\times \frac{34}{19}]\\= 10 [ \frac{126}{19} + 19] \\= 10[ \frac{126 + 361}{19} ]\\= 10 \times \frac{487}{19} \\= \frac{4870}{19}$

Therefore.,

$\red{ Sum \: of \: first \:20 \: terms}\green {=\frac{4870}{19}}$

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