Question

Guys please find this answer.
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Answer 1

Answer:

In this question we need to find A4 i.e equal to

a + 3d

Here a = -2

and d = -2

Therefore a + 3d = -2+3(-2) = -2 - 6 = -8

Please mark this answer as brainliest.

Answer 2

Answer:

$\tt First\:four\: terms\:=\:-2 , -4 , -6 , -8 , -10 , .....$

Explanation:

We know that,

General Term of an AP = a + (n-1)d

where,

a = first term

n = number of terms in AP

d = common difference

Given:

$\tt a_1(first \: term) = - 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\\tt n(number \: of \: terms )\: = \: 4 \: \: \: \: \: \: \: \: \: \: \: \: \:\:\: \\\tt d (common \: difference)\: = \: - 2 \: \:\:\: \:$

$a_2 = a + (2 - 1)d \\ a_2 = a + d \: \: \: \: \: \: \: \: \: \: \: \: \:\: \\ a_2= - 2 - 2\:\:\: \\ a_2 = - 2 -2 \: \: \: \: \: \: \: \: \: \: \: \: \: \\ a_2 = -4 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$a_3 = a + (3 - 1)d \\ a_3 = a + 2d \: \: \: \: \: \: \: \: \: \: \: \: \\ a_3 = - 2 + 2 \times - 2 \\ a_3 = - 2 - 4 \: \: \: \: \: \: \: \: \: \: \: \: \\ a_3= -6 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$a_4 = a + (4 - 1)d \\ a_4 = a + 3d \: \: \: \: \: \: \: \: \: \: \: \: \\ a_4 = - 2 + 3 \times - 2 \\ a_4 = - 2 - 6 \: \: \: \: \: \: \: \: \: \: \: \: \\ a_4 = -8 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

AP = -2 , -4 , -6 , -8 , -10 , -12 , ......

Therefore, the answer is -2 , -4 , -6 and -8.

Other AP Formulas:

nth term of an AP formulas

$\\\\\sf 1) \: n_{th} \: term \: of \: any \: AP \: = a + (n - 1)d$

$\sf 2) \:n_{th} \: term \: from \: the \: end \: of \: an \: AP \: = a + (m - n)d$

$\sf 3) \:n_{th} \: term \: from \: the \: end \: of \: an \: AP = l - (n - 1)d$

$\sf 4) \: Difference \: of \: two \: terms = (m - n)d$

where m and n is the position of the term in AP

$\sf 5) \:Middle\: term\: of\: a\: finite\: AP\:$

$\sf (i) \:\: If \: n \: is \: odd = \frac{n + 1}{2}\:th\:term$

$\sf (ii) \:\: If \: n \: is \: even = \frac{n}{2} \:th \: term \: and \: ( \frac{n}{2} + 1)th \: term$

Sum Formulas

$\sf 1) \: Sum \: of \: first \: n \: terms \: of \: an \:AP = \frac{n}{2} [ \: 2a + (n - 1)d \: ]$

$\sf 2) \: Sum \: of \: first \: n \: natural \: numbers = \frac{n(n + 1)}{2}$

$\sf 3) \: Sum \: of \: AP \: having \: last \: term = \frac{n}{2} [ \: a + l \: ]$