Question

amswer it fast........​

Answer 1

Answer:

It will need 14 terms.. In an A. P.

Answer 2

☯ AnSwEr :

(5).

First term (a) = 12

Sixth term of A.P (a6) = a + 5d = 8 .....(1)

★ Put value of a in equation 1.

→ a + 5d = 8

→ 12 + 5d = 8

→ 5d = 8 - 12

→ 5d = -4

→ d = -4/5

Now, we know that

$\Large{\implies{\boxed{\boxed{\sf{S_n = \frac{n}{2} \bigg(2a + (n - 1)d \bigg)}}}}}$

★ Putting Values ★

$\sf{\dashrightarrow 120 = \frac{n}{2} \bigg(2(12) + (n - 1) \times \frac{-4}{5} \bigg)} \\ \\ \sf{\dashrightarrow 120 \times 2 = n \bigg(24 + (n - 1) \times \frac{-4}{5} \bigg)} \\ \\ \sf{\dashrightarrow 240 = 4n \bigg(6 + (n - 1) \frac{-1}{5} \bigg)} \\ \\ \sf{\dashrightarrow \frac{240}{4} = n( 6 + \frac{-1n}{5} + \frac{1}{5}} \\ \\ \sf{\dashrightarrow 60 = n( 6 + \frac{30 + -n + 1}{5}} \\ \\ \sf{\dashrightarrow 60 = n(\frac{31 - n}{5})} \\ \\ \sf{\dashrightarrow 60 \times 5 = 31n - n^2} \\ \\ \sf{\dashrightarrow -n^2 + 31n - 300 = 0} \\ \\ \sf{\dashrightarrow n^2 - 31n + 300 = 0}$

$\rule{200}{2}$

(6).

A.P : 12, 18, 24 ........ 96

First term (a) = 12

Common Difference (d) = 6

Last term (An) = 96

We know that,

$\Large{\implies{\boxed{\boxed{\sf{A_n = a + (n - 1)d}}}}}$

★ Putting Values ★

$\sf{\dashrightarrow 96 = 12 + (n - 1)6} \\ \\ \sf{\dashrightarrow 96 - 12 = (n - 1)6} \\ \\ \sf{\dashrightarrow \frac{84}{6} = n - 1} \\ \\ \sf{\dashrightarrow n - 1 = 14} \\ \\ \sf{\dashrightarrow n = 15} \\ \\ \Large{\implies{\boxed{\boxed{\sf{n = 15}}}}}$