Question

Insert 6 numbers between 6 and 34 that the
resolving sequence is AP.​

Answer 1

Answer:

if the common difference is 5

then our Ap will be form

7 , 12 , 17 , 22 , 27 , 32 this Ap is formed between 6and 34 when we have to insert 6 number.

Answer 2

$\large\underline{\sf{Solution-}}$

Let the inserted number be

$\rm :\longmapsto\:A_1,A_2,A_3,A_4,A_5,A_6$

so that

$\rm :\longmapsto\:6,A_1,A_2,A_3,A_4,A_5,A_6,34 \: are \: in \: AP$

Now,

↝ First term of an AP, a = 6

↝ Number of terms, n = 8

↝ 8ᵗʰ term of an AP = 34

Let assume that common difference of an AP be 'd'.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

↝ aₙ is the nᵗʰ term.

↝ a is the first term of the sequence.

↝ n is the no. of terms.

↝ d is the common difference.

Tʜᴜs,

$\rm :\longmapsto\:a_8 = a + (8 - 1)d$

$\rm :\longmapsto\:a_8 = a + 7d$

On substituting the values, we get

$\rm :\longmapsto\:34 = 6 + 7d$

$\rm :\longmapsto\:34 - 6 = 7d$

$\rm :\longmapsto\:28 = 7d$

$\bf\implies \:d = 4$

Hence,

$\rm :\longmapsto\:A_1 = a + d = 6 + 4 = 10$

$\rm :\longmapsto\:A_2 = a + 2d = 6 + 8= 14$

$\rm :\longmapsto\:A_3 = a + 3d = 6 + 12= 18$

$\rm :\longmapsto\:A_4 = a + 4d = 6 + 16= 22$

$\rm :\longmapsto\:A_5 = a + 5d = 6 + 20= 26$

$\rm :\longmapsto\:A_6 = a + 6d = 6 + 24= 30$

Additional Information :-

1. If there are n distinct positive numbers, then AM > GM

2. If n numbers are inserted between two numbers a and b such that resulting series is an AP, then n numbers are called Arithmetic mean inserted between a and b.

3. Sum of 'n' arithmetic mean inserted between a and b is n times the single arithmetic mean between a and b.