
If ${t_{11}}$ and ${t_{16}}$ of an AP are 38 and 73 respectively, then ${t_{31}}$ is
A. 178
B. 177
C. 176
D. 175
Answer
553.8k+ views
Hint: We know that the nth term of the AP is given by the formula ${t_n} = a + \left( {n - 1} \right)d$ and using the two given terms in this formula we get two equations and subtracting one from another we get the value of d and using that in any one of the equation we get the value of a. Now with a and d we can find the required value.
Complete step by step solution:
We are given that the 11th term and 16th term of an AP are 38 and 73
We know that the nth term of an AP is given by the formula
$ \Rightarrow {t_n} = a + \left( {n - 1} \right)d$
Where a is the first term and d is the common difference
Now from the given details we get
$ \Rightarrow {t_{11}} = a + \left( {11 - 1} \right)d = a + 10d = 38$………..(1)
$ \Rightarrow {t_{16}} = a + \left( {16 - 1} \right)d = a + 15d = 73$……..(2)
Now let's subtract (1) from (2)
We get
$
\Rightarrow 73 - 38 = a + 15d - a - 10d \\
\Rightarrow 35 = 5d \\
\Rightarrow d = \dfrac{{35}}{5} = 7 \\
$
Now we get the common difference to be 7
Lets substitute this in (1) to get the value of a
$
\Rightarrow a + 10\left( 7 \right) = 38 \\
\Rightarrow a + 70 = 38 \\
\Rightarrow a = 38 - 70 = - 32 \\
$
We get the first term to be – 32
Now we are asked to find the value of ${t_{31}}$
Using the nth term formula we get
$
\Rightarrow {t_{31}} = a + \left( {31 - 1} \right)d \\
\Rightarrow {t_{31}} = a + 30d \\
$
Lets substitute the value of a and d
$
\Rightarrow {t_{31}} = - 32 + 30\left( 7 \right) \\
\Rightarrow {t_{31}} = - 32 + 210 \\
\Rightarrow {t_{31}} = 178 \\
$
Hence we get the required value
Therefore the correct answer is option A.
Note :
In an Arithmetic Sequence the difference between one term and the next is a constant.
We can find the common difference of an AP by finding the difference between any two adjacent terms.
If we know the initial term, the following terms are related to it by repeated addition of the common difference. Thus, the explicit formula is
$\text{Term }= \text{Initial term} + \text{common difference} \times \text{Number of steps from the initial term}$.
Complete step by step solution:
We are given that the 11th term and 16th term of an AP are 38 and 73
We know that the nth term of an AP is given by the formula
$ \Rightarrow {t_n} = a + \left( {n - 1} \right)d$
Where a is the first term and d is the common difference
Now from the given details we get
$ \Rightarrow {t_{11}} = a + \left( {11 - 1} \right)d = a + 10d = 38$………..(1)
$ \Rightarrow {t_{16}} = a + \left( {16 - 1} \right)d = a + 15d = 73$……..(2)
Now let's subtract (1) from (2)
We get
$
\Rightarrow 73 - 38 = a + 15d - a - 10d \\
\Rightarrow 35 = 5d \\
\Rightarrow d = \dfrac{{35}}{5} = 7 \\
$
Now we get the common difference to be 7
Lets substitute this in (1) to get the value of a
$
\Rightarrow a + 10\left( 7 \right) = 38 \\
\Rightarrow a + 70 = 38 \\
\Rightarrow a = 38 - 70 = - 32 \\
$
We get the first term to be – 32
Now we are asked to find the value of ${t_{31}}$
Using the nth term formula we get
$
\Rightarrow {t_{31}} = a + \left( {31 - 1} \right)d \\
\Rightarrow {t_{31}} = a + 30d \\
$
Lets substitute the value of a and d
$
\Rightarrow {t_{31}} = - 32 + 30\left( 7 \right) \\
\Rightarrow {t_{31}} = - 32 + 210 \\
\Rightarrow {t_{31}} = 178 \\
$
Hence we get the required value
Therefore the correct answer is option A.
Note :
In an Arithmetic Sequence the difference between one term and the next is a constant.
We can find the common difference of an AP by finding the difference between any two adjacent terms.
If we know the initial term, the following terms are related to it by repeated addition of the common difference. Thus, the explicit formula is
$\text{Term }= \text{Initial term} + \text{common difference} \times \text{Number of steps from the initial term}$.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

