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Question: If \({t_{11}}\) and \({t_{16}}\) of an AP are 38 and 73 respectively, then \({t_{31}}\) is A. 178 ...

If t11{t_{11}} and t16{t_{16}} of an AP are 38 and 73 respectively, then t31{t_{31}} is
A. 178
B. 177
C. 176
D. 175

Explanation

Solution

We know that the nth term of the AP is given by the formula tn=a+(n−1)d{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
⇒tn=a+(n−1)d\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
⇒t11=a+(11−1)d=a+10d=38\Rightarrow {t_{11}} = a + \left( {11 - 1} \right)d = a + 10d = 38………..(1)
⇒t16=a+(16−1)d=a+15d=73\Rightarrow {t_{16}} = a + \left( {16 - 1} \right)d = a + 15d = 73……..(2)
Now let's subtract (1) from (2)
We get
⇒73−38=a+15d−a−10d ⇒35=5d ⇒d=355=7  \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
⇒a+10(7)=38 ⇒a+70=38 ⇒a=38−70=−32  \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 t31{t_{31}}
Using the nth term formula we get
⇒t31=a+(31−1)d ⇒t31=a+30d  \Rightarrow {t_{31}} = a + \left( {31 - 1} \right)d \\\ \Rightarrow {t_{31}} = a + 30d \\\
Lets substitute the value of a and d
⇒t31=−32+30(7) ⇒t31=−32+210 ⇒t31=178  \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
Term =Initial term+common difference×Number of steps from the initial term\text{Term }= \text{Initial term} + \text{common difference} \times \text{Number of steps from the initial term}.