Question

The sum of 3 A. Ms between A and 22 is 42, then value of A is​

Answer 1

️$\qquad$☀ Let , $\bf A_1$ ,$\bf A_2$,$\bf A_3$ be the three A.M.'s between A and 22 such that A,$\bf A_1$, $\bf A_2$, $\bf A_3$,22 are in A.P.

Where –

• $\sf A = a  $
• $\sf A_1$ = a + d
• $\sf A_3$ = a+2d  
• $\sf A_4$= a + 3d  
• $\sf A_5$ = a + 4d

$\qquad$______________________

We are given –

$\qquad$ $\twoheadrightarrow\bf A_5 \: OR \: a+4d = 22$

$\qquad$ $\pink{\twoheadrightarrow\bf a +4d = 22 ----\:Eq(1)}$

According to the question –

• The sum of 3 A. M.'s between A and 22 is 42.

$\qquad$ $\twoheadrightarrow\bf a + d + a + 2d + a + 3d = 42$

$\qquad$ $\twoheadrightarrow\bf 3a + 6d = 42$

$\qquad$ $\pink{\twoheadrightarrow\bf a + 2d = 14 ----\: Eq(2) }$

Here, we got value of 3rd terms. Now substrate Eq(2) from Eq(1)–

$\qquad$ $\twoheadrightarrow\bf (a + 4d) - (a + 2d) =  22 - 14$

$\qquad$ $\twoheadrightarrow\bf 2d = 8$

$\qquad$ $\purple{\twoheadrightarrow\bf d = 4}$

• Value of d ( Common difference) is 4.

Now,let's put the value of d in Eq(1)–

$\qquad$ $\purple{\twoheadrightarrow\bf a +4d = 22}$

$\qquad$ $\twoheadrightarrow\bf a = 22 -4 \times 4$

$\qquad$ $\twoheadrightarrow\bf a = 22 -16$

$\qquad$ $\purple{\twoheadrightarrow\bf a = 6}$

• Henceforth, value of a (A) is 6.

$\qquad$______________________

Answer 2

The first term is a, hence the value of A is 6

Arithmetic mean

The nth term of an Arithmetic sequence is expressed as:

Tn = a+(n-1)d

where:

• a is the first term
• n is the number of terms
• d is the common difference

Let the Arithmetic between A and 22 to be a, a+d and a+2d

If the sum is 42, hence;

a+d+a+2d+a+3d = 42

3a + 6d = 42

a + 2d = 14....... 1

Also if the 5th term is 42, hence:

a + 4d = 22

From equation 1, a = 14 - 2d

Equation 2 will become 14-2d+4d = 22

2d = 8

d = 4

Since a = 14-2d

a = 14-8

a = 6

Since the first term is a, hence the value of A is 6.

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