Question

The sum of third and seventh term of an A.P. is 6 and their product is 8. Find the sum of the first sixteen terms of A.P.

Answer 1

Hint: We will first assume the first term and the common difference and then form two different equations as per the data given to us. Using that data, we will find the values of a and d. Then, we will put in the values in the formula of sum of terms on an A.P. and we will have the required answer.

Complete step-by-step answer:
Let us first discuss the formula required for A.P.
The \[{n^{th}}\] term on an A.P. is given by the formula:-
${a_n} = a + (n - 1)d$ , where a is the first term of the A.P. and d is the common difference. ……..(1)
The sum of an A.P. up to n terms is given by the formula:-
${s_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right)$, where a is the first term of the A.P. and d is the common difference. ……..(2)
Now, coming back to our question.
Let us assume that the first term of the A.P. given to us be a and the common difference be d.
We are already given that:- The sum of third and seventh term of an A.P. is 6.
Now, let us find ${a_3}$ by putting n = 3 in (1), we will get:
${a_3} = a + (3 - 1)d = a + 2d$ …………(3)
Now, let us find ${a_7}$ by putting n = 7 in (1), we will get:
${a_7} = a + (7 - 1)d = a + 6d$ …………(4)
Adding up the third and seventh term, we will get 6. So, by (3) and (4), we will get:-
${a_3} + {a_7} = a + 2d + a + 6d = 6$
Simplifying it, we will get:-
$ \Rightarrow 2a + 8d = 6$
Taking 2 common from both the sides and cutting it off:-
$ \Rightarrow a + 4d = 3$
$ \Rightarrow a = 3 - 4d$ …………(5)
Now, we are also given that the product of third and seventh term is 8. So, by using (3) and (4), we will get:-
$ \Rightarrow {a_3} \times {a_7} = (a + 2d)(a + 6d) = 8$
Putting (5) in this, we will get:-
$ \Rightarrow (3 - 4d + 2d)(3 - 4d + 6d) = 8$
On simplifying it, we will get:-
$ \Rightarrow (3 - 2d)(3 + 2d) = 8$
Now, we will use the formula:- \[(a + b)(a - b) = {a^2} - {b^2}\].
$ \Rightarrow 9 - 4{d^2} = 8$
Taking 9 from LHS to RHS, we will get:-
$ \Rightarrow - 4{d^2} = - 1$
Cutting off the negative sign and taking 4 from LHS to RHS:-
$ \Rightarrow {d^2} = \dfrac{1}{4}$
$ \Rightarrow d = \pm \dfrac{1}{2}$
Case I: $d = \dfrac{1}{2}$
Putting this in (5), we will get:- $a = 3 - \dfrac{4}{2} = 1$
Now putting these values and n = 16 in (2), we will get:-
$ \Rightarrow {s_{16}} = \dfrac{{16}}{2}\left( {2 + (16 - 1) \times \dfrac{1}{2}} \right) = 76$
Case II: $d = - \dfrac{1}{2}$
Putting this in (5), we will get:- $a = 3 + \dfrac{4}{2} = 5$
Now putting these values and n = 16 in (2), we will get:-
$ \Rightarrow {s_{16}} = \dfrac{{16}}{2}\left( {2 \times 5 + (16 - 1) \times \left( { - \dfrac{1}{2}} \right)} \right) = 20$

Hence, the sum up to 16 terms is either 76 or 20.

Note: The students might make the mistake of considering only one value of d that is positive but then their answer will only be partially right because they have completely forgotten the other possibility.
The students must note that we formed two equations for two unknown variables a and d. We need as many equations as many numbers of unknown variables we have.