Question

The sum of the first five terms of the AP: 3, 7, 11, 15,...... is

Answer 1

Answer: ............down ↓↓↓↓↓ ...................

Step-by-step explanation:

Sn = n / 2 * [ 2a + ( n - 1 ) * d ]

Sn= 5 / 2 * [ 2* 3 + ( 5 - 1 ) * 4 ]

Sn = 5/2 * [ 6 + 6 * 4 ]

Sn = 5/2 * [ 6+ 24 ]

Sn = 5/2 * 30

Sn = 75

Answer 2

$Given \: A.P : 3,7,11,15,\ldots$

$First \: terms (a = a_{1} ) = 3$

$Common \: difference (d) = a_{2} - a_{1}$

$= 7 - 3$

$= 4$

/* We know that , */

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

$Here , a = 3 , d = 4 \: and \: n = 5$

$\red{ Sum \: of \: first \: five \: terms }$

$= \frac{5}{2} [ 2\times 3 + ( 5 - 1 )\times 4 ]$

$= \frac{5}{2} ( 6 + 4\times 4 )$

$= \frac{5}{2} \times 22$

$= 5 \times 11$

$= 55$

Therefore.,

$\red{ Sum \: of \: first \: five \: terms }\green {= 55}$

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