The answer
Arithmetic Progression (A.P.) is the subject of this inquiry.
Each term in an arithmetic progression (A.P.) is equal to the term before it plus a constant value known as the common difference. Stated differently, the next term is obtained by adding each term by the common difference.
The first term (a1) in this instance is 3, and the last term (an) is 27. In order to create an arithmetic progression, we are to insert two arithmetic means—which will become new terms in the sequence—between them.
Identifying the shared distinction
Every term in an arithmetic progression is equal to the term before it plus the common difference. Let us use d to represent the common difference. Two equations can be written down using this relationship:
Third term (a2) = a1 + d = 3 + d
“A3” is the third term (a2+d=(3+d)+d=3+2d.
a4 (the fourth term) = a3 + d =(3+2d)+d=3+3d
Additionally, we are informed that the fourth term (a4) equals 27. Let us replace the fourth term in the equation with this value:
3 + 3d = 27
3d=27−3
3d = 24
d = 324
d=8
We now understand that eight is the common difference.
Calculating the mathematical means
The sequence’s second and third terms are the arithmetic means that we must add. With the help of the common difference we just computed, we can locate them.
2nd term (a2) = a1 + d = 3 + 8 = 11.
3rd term (a3) = a2 + d = 11 + 8 = 19
Thus, 11 and 19 are the two arithmetic means that must be inserted between 3 and 27.
The right response is (c) 7,19.
The answer
Arithmetic Progression (A.P.) is the subject of this inquiry.
Each term in an arithmetic progression (A.P.) is equal to the term before it plus a constant value known as the common difference. Stated differently, the next term is obtained by adding each term by the common difference.
The first term (a1) in this instance is 3, and the last term (an) is 27. In order to create an arithmetic progression, we are to insert two arithmetic means—which will become new terms in the sequence—between them.
Identifying the shared distinction
Every term in an arithmetic progression is equal to the term before it plus the common difference. Let us use d to represent the common difference. Two equations can be written down using this relationship:
Third term (a2) = a1 + d = 3 + d
“A3” is the third term (a2+d=(3+d)+d=3+2d.
a4 (the fourth term) = a3 + d =(3+2d)+d=3+3d
Additionally, we are informed that the fourth term (a4) equals 27. Let us replace the fourth term in the equation with this value:
3 + 3d = 27
3d=27−3
3d = 24
d = 324
d=8
We now understand that eight is the common difference.
Calculating the mathematical means
The sequence’s second and third terms are the arithmetic means that we must add. With the help of the common difference we just computed, we can locate them.
2nd term (a2) = a1 + d = 3 + 8 = 11.
3rd term (a3) = a2 + d = 11 + 8 = 19
Thus, 11 and 19 are the two arithmetic means that must be inserted between 3 and 27.
The right response is (c) 7,19.