Question

Answer it plz .........​

Answer 1

☯ 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 ★

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

(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 ★

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

Answer 2

Answer:

5)-----

Let a and d are first term and

common difference of an A.P

According to the problem given,

a1 = a = 12----( 1 )

a6 = 8

a + 5d = 8 ---( 2 )

Put ( 1 ) in equation ( 2 ) ,

12 + 5d = 8

5d = 8 - 12

5d = -4

d = -4/5

Now ,

a = 12 , d = -4/5,

Let the number of terms = n

Sn = 120

n/2 [ 2a + ( n -1 )d ] = 120

n/2 [ 2×12 + ( n - 1 )( -4/5 ) ] = 120

4n/2 [ 6 + ( n - 1 ) ( -1/5 ) ] = 120

2n [ 30 -n + 1 ]/5 = 120

n ( 31 - n ) = 120 × ( 5/2 )

31n - n² = 300

n² - 31n + 300 = 0

Plz , verify the question again there is

error in the data.

6))----

First two digit number=10

Last two digit number=99

First two digit number divisible by 6=12

Last two digit number divisible by 6=96

Now the series is:

12,18,24.................................96

18-12=6

24-18=6

This series clearly forms an AP with first term 12 , common difference 6 and last term=96.

a=12

d=6

anth term=96

a+(n-1)d=96

12+(n-1)6=96

6n-6=84

6n=90

n=15

Hence there are 15 two digits number divisible by 6.