Question

How many terms are in the ap 41 38 35 ..82

Answer 1

Answer:

The Formula is:

l = a+ (n-1) d

8= 41 + (n-1) -3

8= 41 -3n + 3

8 - 44 = -3n

-36 = -3n

n = -36/-3

n = 12.

Step-by-step explanation:

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Answer 2

Given:

$a_1(first \: term) = 41$

$a_l(last \: term) = - 82$

$AP = 41 \: , \: 38 \: , \: 35 \: , \: ...... \: , \: - 82$

To Find:

Number of Terms in the given AP.

Answer:

Here,

$a_1 = 41$

$d = 38 - 41$

$d = - 3$

Explanation:

We know that,

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

$n_l = \: a + (n - 1)d$

Substituting the value of a, n and d, we get:

$n_l = 41 + (n - 1)( - 3)$

$- 82 = 41 - 3(n - 1)$

$- 82 = 41 - 3n + 3$

$- 82 = - 3n + 44$

Sending 44 to RHS, we get:

$- 3n = - 126$

Dividing both sides by -3, we get:

$\frac{ - 3n}{ - 3} = \frac{ - 126}{ - 3}$

$n = 42$

$\therefore \: Number \: of \: terms \: in \: AP \: = 42$

Other AP Formulas:

nth term of an AP formulas

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

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

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

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

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

Middle term of a finite AP

$If \: n \: is \: odd = \frac{n + 1}{2}$

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

Sum Formulas

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

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

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