Question

2 First four terms of abd Al are.
whose first terms is - 2 and common difference
is-2
2​

Answer 1

nth term (n > 1) of AP is defined by a + (n - 1)d, where a is first term and d is common difference.

Therefore, if a is - 2 and d = - 2.

1st term = a = - 2 (given)

For 2nd term, subtitute n = 2, in a + (n - 1)d.

2nd term = a + (2 - 1)d = a + d = -2 + (-2)

2nd term = - 4

For 3rd term, n = 3

3rd term = a + (3 - 1)d = a + 2d

3rd term = -2 + 2(-2) = - 6

For 4th term, n = 4

3rd term = a + (4 - 1)d = a + 3d

3rd term = -2 + 3(-2) = - 8

First four terms are - 2, - 4, - 6, - 8.

Answer 2

Answer:

Appropriate Question :-

• The first four terms of an AP are whose first term is - 2 and common difference is - 2.

Given :-

• The first term of an AP is - 2 and the common difference is - 2.

To Find :-

• What are the first four terms of an AP.

Formula Used :-

$\clubsuit$ General term (nth term) of an AP Formula :

$\longmapsto \sf\boxed{\bold{\pink{a_n =\: a + (n - 1)d}}}$

where,

• n = Number of terms
• a = First term of an AP
• d = Common difference of an AP

Solution :-

Given :

$\bigstar\: \: \rm{First\: term\: of\: an\: AP\: (a)} =\: - 2$

$\bigstar\: \: \rm{Common\: difference \: (d)}=\: - 2$

${\normalsize{\bold{\purple{\underline{\mapsto\: Second\: term\: of\: AP\: :-}}}}}$

Given :

• First term of an AP (a) = - 2
• Number of terms (n) = 2
• Common difference (d) = - 2

According to the question by using the formula we get,

$\implies \sf a_2 =\: - 2 + (2 - 1) \times (- 2)$

$\implies \sf a_2 =\: - 2 + 1 \times (- 2)$

$\implies \sf a_2 =\: - 2 + (- 2)$

$\implies \sf a_2 =\: - 2 - 2$

$\implies \sf \bold{\red{a_2 =\: - 4}}$

${\normalsize{\bold{\purple{\underline{\mapsto\: Third\: term\: of\: AP\: :-}}}}}$

Given :

• First term of an AP (a) = - 2
• Number of terms (n) = 3
• Common difference (d) = - 2

According to the question by using the formula we get,

$\implies \sf a_3 =\: - 2 + (3 - 1) \times (- 2)$

$\implies \sf a_3 =\: - 2 + 2 \times (- 2)$

$\implies \sf a_3 =\: - 2 + (- 4)$

$\implies \sf a_3 =\: - 2 - 4$

$\implies \sf\bold{\red{a_3 =\: - 6}}$

${\normalsize{\bold{\purple{\underline{\mapsto\: Fourth\: term\: of\: AP\: :-}}}}}$

Given :

• First term of an AP (a) = - 2
• Number of term (n) = 4
• Common difference (d) = - 2

According to the question by using the formula we,

$\implies \sf a_4 =\: - 2 + (4 - 1) \times (- 2)$

$\implies \sf a_4 =\: - 2 + 3 \times (- 2)$

$\implies \sf a_4 =\: - 2 + (- 6)$

$\implies \sf a_4 =\: - 2 - 6$

$\implies \sf\bold{\red{a_4 =\: - 8}}$

$\sf\bold{\therefore\: The\: first\: four\: terms\: of\: an\: AP\: is\: - 2, - 4, - 6, - 8\: respectively\: .}$

EXTRA FORMULA :-

$\leadsto \rm{Sum\: of\: first\: n^{th}\: term\: of\: AP\: :- }$

$\mapsto \sf\boxed{\bold{\pink{S_n =\: \dfrac{n}{2}\bigg\lgroup 2a + (n - 1)d\bigg\rgroup}}}$

where,

• a = First term of an AP
• n = Number of terms
• d = Common difference