Question

9.In an A.P if a=5 ,nth term is 81 and sum of n terms of an A.P is 860 then n is *​

Answer 1

hope this helps uh dea....

Answer 2

Given:

a = 5

nth term = 81

Sum of n terms = 860

Answer:

Explanation:

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

$81 = 5 + (n - 1)d$

$(n - 1)d = 76$

$(n - 1) = \frac{76}{d}$

$n = \frac{76}{d} + 1$

Taking LCM, we get:

$\\n = \frac{76 + d}{d} \: ......(1)$

$Sum \: of \: n \: terms \: = \: \frac{n}{2} \left(2a + (n - 1)d\right)$

Substituting the values , we get:

$\\860 \times 2 = n(10 + (n - 1)d) \\ 1720 = n(10 + (n - 1)d) \: \: \: \: \: \:$

Substituting the value of n from equation (1), we get:

$\\1720 = \frac{76 + d}{d} \left(10 + \left(\frac{76 + d}{d} - 1 \right)d\right) \\ \\ 1720 = \frac{76 + d}{d} \left(10 +\left( \frac{76 + d - d}{d}\right)d\right) \\ \\ 1720 = \frac{76 + d}{d} \left(10 + \frac{76}{d} \times d\right) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ 1720 = \frac{76 + d}{d} \times 86 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Sending 86 to LHS,

$\\ \frac{1720}{86} = \frac{76 + d}{d} \\ \\ 20 = \frac{76 + d}{d} \: \: \: \: \:$

Sending d to LHS, we get:

$\\20d = 76 + d \\ 19d \: = \: 76 \: \: \: \: \: \: \: \\ d = 4 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Substituting the value of d in equation (1), we get:

$\\n = \frac{76 + 4}{4} \\ n = \frac{80}{4} \: \: \: \: \: \: \: \\ n = 20 \: \: \: \: \: \: \: \: \:$

⭐More About AP⭐

nth term of an AP formulas

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

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

$\sf 2) \: (ii) \: n_{th} \: term \: from \: the \: end \: of \: an \: AP = l - (n - 1)d$

$\sf 3) \: Difference \: of \: two \: terms = (m - n)d$

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

$\tt 4) \: Middle \: term \: of \: a \: finite \: AP$

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

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

Sum Formulas

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

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

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

⭐Numbers in an AP⭐

Three Numbers in an AP:

(a-d) , a and (a+d)

Four Numbers in an AP:

(a-3d) , (a-d) , (a+d) and (a+3d)

Five Numbers in an AP:

(a-2d) , (a-d) , a , (a+d) and (a+2d)