P.S - Here we will be applying the formula, that is used to find the nth term of an A.P by taking n as 11 or 16 and 31. Then we will find the value of first term and the common difference of the A.P (the most important values to be known) using the substitution method.
The number of series (progression) in which the difference of the successive number is same is termed as the arithmetic progression.
tn = a + (n-1) d
where,
tn - first term
nth - number of terms
To solve ANY question with the help of substitution method the following are the steps -
Step 1 - Find the value of one variable, say x in terms of the other variable, i.e, x from either equation, whichever is convenient.
Step 2 - Substitute this value of y in the other equation, and seduce it to an equation in one variable, i.e, in terms of x, which can be solved sometimes, you can get statements with no variable. If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions of the statement is false, then the pair of linear equations is inconsistent.
Step 3 - Substitute the value of x (or y) obtained in Step 2 in the equation read in Step 1 to obtain the value of the other variable.
✤Solution -
We know, a + 10d = a11 and a + 15d = a16 and as we are provided with the values of a11 and a16. We can also write them as,
a + 10d = 38 --- (1)
a + 15d = 78 --- (2)
Now proceeding with the substitution method,
We can also write equation (1) as,
a = 38 - 10d
Putting the value of a in equation (2)
➠ 38 - 10d + 15d = 78
➠ 5d = 78 - 38
➠ 5d = 40
➠ d = 40/5
➠ d = 8
Putting the value of d in equation (1) to find the value of a,
➠ a + 10(8) = 38
➠ a + 80 = 38
➠ a = 38 - 80
➠ a = -42
Substituting the value of a and d in
a + 30d = a31.
➠ a31 = -42 + 30(8)
➠ a31 = -42 + 240
➠ a31 = 198
Therefore, the value of the 31st termof an A.P is 198.
Answers & Comments
Answer:
Given that, a
11
=38 a
16
=73
We know that,
a
n
=a+(n−1)d
a
11
=a+(11−1)d
38=a+10d ... (i)
Similarly,
a
16
=a+(16−1)d
73=a+15d ... (ii)
On subtracting (i) from (ii), we get
35=5d
∴d=7
From equation (i),
38=a+(10)(7)
⇒38−70=a
∴a=−32
Now a
31
=a+(31−1)d
=−32+30(7)
=−32+210
=178
Hence, 31
st
term is 178.
Verified answer
✤Given -
➢ the 11th term of an A.P is 38.
i.e, a11 = 38
➢ the 16th term of the same A.P is 78.
i.e, a16 = 78
✤To find -
➢ the 31st term of an A.P
✤Concept -
P.S - Here we will be applying the formula, that is used to find the nth term of an A.P by taking n as 11 or 16 and 31. Then we will find the value of first term and the common difference of the A.P (the most important values to be known) using the substitution method.
The number of series (progression) in which the difference of the successive number is same is termed as the arithmetic progression.
tn = a + (n-1) d
where,
To solve ANY question with the help of substitution method the following are the steps -
Step 1 - Find the value of one variable, say x in terms of the other variable, i.e, x from either equation, whichever is convenient.
Step 2 - Substitute this value of y in the other equation, and seduce it to an equation in one variable, i.e, in terms of x, which can be solved sometimes, you can get statements with no variable. If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions of the statement is false, then the pair of linear equations is inconsistent.
Step 3 - Substitute the value of x (or y) obtained in Step 2 in the equation read in Step 1 to obtain the value of the other variable.
✤Solution -
We know, a + 10d = a11 and a + 15d = a16 and as we are provided with the values of a11 and a16. We can also write them as,
a + 10d = 38 --- (1)
a + 15d = 78 --- (2)
Now proceeding with the substitution method,
We can also write equation (1) as,
a = 38 - 10d
Putting the value of a in equation (2)
➠ 38 - 10d + 15d = 78
➠ 5d = 78 - 38
➠ 5d = 40
➠ d = 40/5
➠ d = 8
Putting the value of d in equation (1) to find the value of a,
➠ a + 10(8) = 38
➠ a + 80 = 38
➠ a = 38 - 80
➠ a = -42
Substituting the value of a and d in
a + 30d = a31.
➠ a31 = -42 + 30(8)
➠ a31 = -42 + 240
➠ a31 = 198
Therefore, the value of the 31st term of an A.P is 198.