Question

Find the sum 4+9+14+.......+199

Answer 1

Answer:

$\red { 4+9+14+ \codt\cdot \cdot +199}\green {= 4060}$

Step-by-step explanation:

$Given \: sequence \: 4,9,14,\cdot\cdot\cdot+199$

$First \:term (a) = 4$

$n^{th} \: term = a_{n} = 199\: (given)$

$a_{2} - a_{1} = 9 - 4 = \pink {5}$

$a_{3} - a_{2} = 14 - 9 = \pink {5}$

$[tex] a_{2} - a_{1} = a_{3} - a_{2} = \pink {5}$

$common \: difference (d) = \pink {5}$

$Given \: sequence \: is \: an \: A.P$

$n^{th} \: term = a_{n} = 199\: (given)$

$\implies a + (n - 1)d = 199$

$\implies 4+(n-1)5 = 199$

$\implies (n-1)5 = 199-4$

$\implies (n-1)5 = 195$

$\implies n-1 = \frac{195}{5}$

$\implies n = 39 + 1$

$\implies n = 40$

$Now, 4+9+\cdot \:cdot \: cdot + 199\= Sum \: of \: 40 \: terms \\= S_{40}$

$\boxed { \pink { S_{n} = \frac{n}{2}(a+a_{n})}$

$\implies S_{40} = \frac{40}{2} (4+199)$

$= 20 \times 203$

$\green {= 4060}$

Therefore.,

$\red { 4+9+14+ \codt\cdot \cdot +199}\green {= 4060}$

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