Question

# The $5th$ Term of an A.P. is thrice the $2nd$ term and the $12th$ term exceeds the $6th$ term by $1$ . Find the $16nth$ Term.

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Hint: After having a look we can identify that this is a sum from the Arithmetic Progression. First step in solving such a sum is to understand each and every statement and then think of the correct formula which can be applied. Based on the given data we can apply the formula ${T_n} = a + (n - 1)d$ to find the ${N^{th}}$ term. In order to solve these problems we will have to pick up 1 statement at a time and then form equations accordingly.

Since this sum involves finding the value of ${N^{th}}$ term , we will have to use the formula ${T_n} = a + (n - 1)d$ , where
${T_n}$ - Value of ${N^{th}}$ term
$a$ - $1st$ term of the given series
$n$ - No. of terms in the Series
$d$ - common difference between $2$ consecutive terms in the series
Considering the first statement of the numerical, we can form the following equations
${T_5} = a + 4d...............(1)$
${T_2} = a + d.................(2)$
Given that ${T_5} = 3 \times {T_2}$
From $Equation1\& Equation2$ we can form the following relation
$\Rightarrow a + 4d = 3 \times (a + d)...........(3)$
$\Rightarrow a + 4d = 3a + 3d...........(4)$
$\Rightarrow d = 2a...........(5)$
Considering second statement in the numerical ,we can form following equations
${T_{12}} = a + 11d............(6)$
${T_6} = a + 5d............(7)$
It is given that ${T_{12}} - {T_6} = 1.............(8)$
Substituting values of $Equation6\ & \ Equation7$ in $Equation8$ we get the following equation
$\Rightarrow (a + 11d) - (a + 5d) = 1$
$\Rightarrow 11d - 5d = 1$
$\Rightarrow d = \dfrac{1}{6}$
We can use $Equation5$ to find the value of $a$ .
$\Rightarrow a = \dfrac{1}{6} \times 2 = \dfrac{1}{3}$
Since we have the value of $a\& d$ we can now find the $16th$ term
$\Rightarrow {T_{16}} = a + 15d$
Substituting values of $a\& d$ we get the final value of $16th$ term
$\Rightarrow {T_{16}} = \dfrac{1}{3} + 15 \times \dfrac{1}{6}$
In order to simplify the sum we can multiply the numerator and denominator of $\dfrac{1}{3}$ by $2$ , so that we have a common denominator.
$\Rightarrow {T_{16}} = \dfrac{2}{6} + \dfrac{{15}}{6}$
$\Rightarrow {T_{16}} = \dfrac{{17}}{6}$
Thus the final Answer is ${T_{16}} = \dfrac{{17}}{6}$
So, the correct answer is “${T_{16}} = \dfrac{{17}}{6}$”.

Note: Though this sum may look complicated or lengthy , it is not at all difficult as we have applied only 1 formula several times . In numericals related to arithmetic progression it is advantageous to learn the formula of Sum upto ${N^{th}}$ term also . These are the only 2 formulas based on which the questions are asked. Students are advised not to panic if the question is long or difficult to comprehend, it will become pretty simple if they break down multiple statements into smaller ones and start solving.